Question

A golf ball is struck at ground level. The speed of the golf ball as a function of the time is shown in Figure 4-36, where t = 0 at the instant the ball is struck. The scaling on the vertical axis is set by va=16 m/s and vb=32 m/s. Answer the following questions:

(a) How far does the golf ball travel horizontally before returning to ground level?
(b) What is the maximum height above ground level attained by the ball?

Answer 1

Step-by-step explanation:

initial total speed of the ball is given as

`v_b = 32 m/s`

minimum speed during its trajectory

`v_a = 16 m/s`

now the speed in x direction will be the minimum speed

`v_x = 16 m/s`

also we know that

`v_y^2 + v_x^2 = 32^2`

`v_y^2 + 16^2 = 32^2`

`v_y = 27.7 m/s`

now total time of the flight

`T = (2v_y)/(g)`

`T = (2(27.7))/(9.8) = 5.65 s`

now horizontal range will be

`R = 16(5.65) = 90.5 m`

Part b)

as we know that at maximum height vertical velocity is zero

so we will have

`v_(fy)^2 - v_(iy)^2 = 2as`

now we have

`0 - 27.7^2 = 2(-9.8)(H)`

`H = 39.15 m`