Question

A golf ball is struck at ground level. The speed of the golf ball as a function of the time is shown in the figure, 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= 29.3 m/s. (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?

by the way, 16 m/s is the minimum value on this graph and 29.3 is the maximum value on this graph. 29.3 appears first and then 16 is in the middle in a concave-up parabola. So this is what I did and I wonder if this is right:
So the initial velocity, a mixture of both x and y components is=29.3. To find the theta, we can take 29.3cos(theta)=16. Why do we use 16? Because the minimum speed corresponds to a vertical velocity of 0, meaning 16 is the constant horizontal velocity. So then, theta is =56.91 degrees. We then have to solve for initial y component of velocity, so we have 29.3sin(56.91)=24.55. We then say, since the graph has 5 seconds at the end point: s=.5(16+16)*5=80m. Is that right? for a. And then is b just s=.5(24.546+0)*2.5=30.6825? Thanks for the help, I just am unsure. Thanks and have great day :)

Answer 1

time of flight T = 4 sec
time to reach max height = 2 s
=====================
if (u = 34 m/s) was initial speed and (p) angle of projection
u cos (p) *T = Horizontal Range
R = 4 u cos (p) ---------- (1)
--------------------------------
y = u sin (p) * t - 0.5 gt^2 ------- (2)
vy = speed vertical compo =dy/dt= u sin (p) - gt
vy =0 at max height
t (max) = 2 =given = u sin (p) /g
u sin (p) = 2g ------------------------------------ (3)
sin (p) = 2*9.8/34 = 0.5765
so > cos (p) = 0.8171
Range = R = 4 u cos (p) = 4*34*0.8171
R = 111.13 meter
==================
max height
y (max) = u sin (p) * 2 - 0.5 g*2^2
y (max) = 2g * 2 - 0.5 g*4
y (max) = 4g - 2g = 2g = 2*9.8 = 19.6 meter