-4 points SerCP11 2.P 045 My Notes Ask Your A ball is thrown vertically upward w
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-4 points SerCP11 2.P 045 My Notes Ask Your A ball is thrown vertically upward with a speed of 26.0 m/s. (a) How high does it rise? (b) How long does it take to reach its highest point? (c) How long does the baltake to hit the ground after it reaches its highest point? (d) What is its velocity when it returns to the level from which it started? m/s Need Help? Submit ArswerSaveProrss Pracce Another Version 6 points SerCP11 2 P051 A tennis player tosses a tennis ball straight up and then catches it after 1.66 s at the same height as the point of release My Notes Ask Your (a) What is the acceleration of the ball while it is in flight? m/s2 direction (b) What is the velocity of the ball when it reaches its maximum height? m/s direction (c) Find the initial velocity of the ball m/s upward (d) Find the maximum height it reaches. Need Help?Explanation / Answer
1)
As the ball rises its speed decreases at a rate equal to 9.8 m/s^2 which is the gravitational acceleration acting in the downward direction.
The ball rises till its speed becomes 0. If the ball rises for t seconds.
0 = 26.0 - 9.8*t
t = 26.0/9.8 = 2.65 seconds
The height of the ball after 2.65 seconds is 26.0*2.65 - 1/2*9.8*2.65^2 = 34.5 m
The time taken by the ball to return to the ground is also 2.65 seconds.
Its velocity when it returns is 26.0 m/s in the downward direction.
2)
a) Acceleration = 9.8 m/s^2
b) Velocity at maximum height = 0
c) Using the general kinematic equation: Y = Yo + Voy t - (1/2)gt^2
Substituting: 0 = 0 + Voy t - 4.9 t^2
0 = Voy (1.66) - 4.9(1.66)^2
Voy = 8.134 m/s
d)
Using the kinematic equation: V^2 = Vo^2 -2gd
Substituting: (0)^2 = (8.134)^2 - 2(9.8)d
d = 3.38 m
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