Concept explainers
How long after the ball is released from the balcony, the friend has to wait, to start running so that she'll be able to catch the ball exactly 1.00 m above the floor of the court.
Answer to Problem 63QAP
The friend has to wait for 0.25 s after the ball is released and then start running so as to catch the ball at the height of 1.00 m above the floor.
Explanation of Solution
Given info:
The height of the balcony above the court
Initial velocity of the ball
Angle at which the ball is released
Distance of the initial position of the friend from the balcony
Friend's initial velocity
Friend's acceleration
Height at which the ball is caught
Formula used:
The equations of motion for vertical and horizontal motion of the ball can be used to find the time the friend needs to wait.
For vertical motion,
Here,
For horizontal motion of the ball,
Here,
The friend's motion can be analyzed using the equation,
Here,
Calculation:
Assume the origin A to be located at the point just below the balcony, with the x axis parallel to the ground and the positive y axis directed upwards. The height of the balcony from the ground is OA. The friend stands at B initially, and then catches the ball at the point C at a height CD from the ground. This is represented by the diagram shown below.
The ball is released with an initial velocity
The vertical motion of the ball is governed by the gravitational force. The acceleration of the ball in the vertical direction is equal to the acceleration of free fall.
Therefore,
The ball makes a vertical displacement from the initial position
Therefore,
Rewrite the equation (1) using the above expression, equation (4).
Substitute the values of the variables in the equation and calculate the value of time of flight t.
Rewrite the equation as a quadratic for t.
Solve for t.
Take the positive root alone.
The ball is in flight for 1.71 s. During this time, the ball travels a horizontal distance
Therefore,
Since point A is directly below the balcony, its x coordinate is
Therefore,
Use equation (5) and the values of the variables in the equation (2) and calculate the value of x.
The friend stands at the position
The horizontal displacement the friend needs to make is given by,
In equation (3), substitute the known values of the variables and calculate the time
Simplify the expression and solve for
The ball takes a time
Calculate the time
Conclusion:
Thus, the friend has to wait for 0.25 s after the ball is released and then start running so as to catch the ball at the height of 1.00 m above the floor.
Want to see more full solutions like this?
Chapter 3 Solutions
COLLEGE PHYSICS
- Question 1:A soccer player kicks a ball with an initial velocity of 20 m/s at an angle of 30 degrees above the horizontal. Calculate the following: a) The time it takes for the ball to reach its maximum height.b) The maximum height the ball reaches above its starting point.c) The total time the ball is in the air.d) The horizontal distance the ball travels before hitting the ground. NOTE I need the answer for the full question thank youarrow_forward• A glaucous-winged gull, ascending straight upward at 5.20 m>s,drops a shell when it is 12.5 m above the ground. (a) What arethe magnitude and direction of the shell’s acceleration just afterit is released? (b) Find the maximum height above the groundreached by the shell. (c) How much time does it take for the shellto reach the ground? (d) What is the speed of the shell just beforeit hits the ground?arrow_forwardA particle initially located at the origin has an acceleration of a=3.00jm/s2 and an initial velocity of vi=5.00im/s. Find (a) the vector position of the particle at any time t, (b) the velocity of the particle at any time t, (c) the coordinates of the particle at t = 2.00 s, and (d) the speed of the particle at t = 2.00 s.arrow_forward
- A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long as 2.2 m have been recorded. If a frog jumps 2.2 m and the launch angle is 45, find (a) the frogs launch speed and (b) the time the frog spends in the air. Ignore air resistance.arrow_forwardA projectile is launched from the point (x = 0, y = 0), with velocity (12.0i19.0j)m/s2, at t = 0. (a) Make a table listing the projectiles distance r from the origin at the end of each second thereafter, for 0 t 10 s. Tabulating the x and y coordinates and the components of velocity vx and vy will also be useful. (b) Notice that the projectiles distance from its starting point increases with time, goes through a maximum, and starts to decrease. Prove that the distance is a maximum when the position vector is perpendicular to the velocity. Suggestion: Argue that if v is not perpendicular to r, then r must lie increasing or decreasing. (c) Determine the magnitude of the maximum displacement. (d) Explain your method for solving part (c).arrow_forwardA device launches horizontally two identical balls (A and B) simultaneously (at the same instant of time) from the same height h = 5m. If the initial speed of ball A is vA = 2m/s and the initial speed of ball B is vig = 4m/s, then: (g 10m/s) Launch device A Both balls (A and B) would hit the ground at the same time t = 1s. B Both balls (A and B) would hit the ground at the same time t= 2s. C Both balls (A and B) wvould hit the ground at the same time t= 15s. D. Ball A would hit the ground after time t= Is and ball B would hit the ground after time t = 2s. E Ball A would hit the ground after time t 2s and ball B would hit the ground after time t 1s. A ball is kicked from the ground at point O with an intial velocity 1 = 20 m/s that makes an agle 0= 30° with the horizontal This ball reaches its maximum height at pouLA and then returns to the ground at point B as shown in the figure (g 10m/s") The tine taken by the ball to reach the maximurm height at point A (toA) is equal to:arrow_forward
- You toss a ball horizontally at 9 m/s about 1.2 m above the ground towards a dog who is sitting 3.5 m away. • What height will the dog’s mouth need to be at to catch the ball (if the dog only jumps vertically)?• How fast (and in what direction) will the ball be traveling when the dog catches it?• Include a sketch of the situation (with a coordinate system).arrow_forwardA swimmer runs horizontally off a diving board with aspeed of 3.32 m/s and hits the water a horizontal distance of1.78 m from the end of the board. (a) How high above the waterwas the diving board? (b) If the swimmer runs off the board with a reduced speed, does it take more, less, or the same time to reach the water? Graph and Explain.arrow_forward• A ball is thrown straight up from the ground with speed v0. At the same instant, a second ball is dropped from rest from a height H, directly above the point where the first ball was thrown upward. There is no air resistance. (a) Find the time at which the two balls collide. (b) Find the value of H in terms of v0 and g so that at the instant when the balls collide, the first ball is atarrow_forward
- A projectile is launched from ground level at an angle of 30 degrees above the horizontal. Neglect air resistance and consider the motion from just after the moment it is launched to just before the moment it lands on the ground. When is the projectile’s acceleration equal to zero? a) Just after launch. b) Halfway to the highest point. c) Just before landing on the ground. d) At the highest point. e)The projectile's acceleration is never zero. f) The projectile's acceleration is always zero. At what point do the velocity and the acceleration have the same direction? a) Just after launch. b) Halfway to the highest point. c) Just before landing on the ground. d) At the highest point. e) The projectile’s velocity and acceleration always have the same direction. f) The projectile’s velocity and acceleration never have the same direction. At what point are the velocity and…arrow_forwarda) What is the distance it travelled during 2 seconds?b) In what direction did it ravel (angle with the positive x-axis)? c) What is the acceleration vector of this particle?arrow_forwardProblem 5: A ball is thrown horizontally from the top of a 70 m building and it lands 90 m from the base of the building. a) How long is the ball in the air? b) What must have been the initial speed of the ball? c) What is the vertical component of the ball's velocity just before it hits the ground? d) What is the ball's speed just before it hits the ground?arrow_forward
- Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning