A ball is thrown from a top with an initial downward velocity of magnitude V_0 =

Question

A ball is thrown from a top with an initial downward velocity of magnitude V_0 = 2.7 m/s. The top is a distance above the ground, h = 43 m. In this problem use a coordinates system in which upwards is positive. a) Find the vertical component of the velocity, V_f, in meters per second, with which the ball hits the ground. b) If we wanted the ball's final speed to be exactly 27.3 m/s from what height, h_new (in meters) would be to throw it with the same initial velocity? c) If the height is fixed at 43 m, but what we wanted the ball's final speed to be 32.7 m/s, what would the vertical component of the initial velocity V_iy, need to be, in meters per second?

Explanation / Answer

a) The initial vertical velocity of the ball is
v0   = 2.7 m/s
Using Newton's equations of motion
v2  = v02  + 2 a S
Where S is the distance and a is the vertical acceleration
Here a = -g
v2  = v02 - 2 g S
v2  = (2.7)2 - 2 x 9.81 x (- 43m)
v2  = 850.95
v = 29.17 m/s
This is the final velocity
b) Let H be the new height, so that the balls final velocity is 27.3 m/s.
v2  = v02 + 2 g H
H = v2- v02 / 2 g
H = 27.32  - 2.72 / (2 x 9.81)
H = 37.61 m
The new height should be 37.61 m
c) Final speed is 32.7 m and the height is 43m
v2  = v02 + 2 g H
v02  = v2  - 2 g H
v02  = 32.72  - 2 x 9.81 x 43
v02 = 225.63
v0  = 15.02 m/s

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