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

Answer:

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

$T = \frac{2v_y}{g}$

$T = \frac{2(27.7)}{9.8} = 5.65 s$

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

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

$H = 39.15 m$

Answer 2

In your question where as a golf ball is struck at a ground level and the speed of the ball as a function of time is in the figure where time t=0 and va = 16m/s and vb=32m/s. The following is the answer: 
a) How far does the golf ball travel horizontally before returning to ground level? 
-80m
(b) What is the maximum height above ground level attained by the ball?
-39.87m