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. (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?

Question

anonymous · Answer

attachment

anonymous · Answer

This is tricky since we don't know the components of the velocity. 

Let's note from our equations of motion that the range of the projectile is$$r = {v^2 \sin(2 \theta) \over g}$$the maximum height is$$y_{\max} = {v^2 \sin^2(\theta) \over 2g}$$and the time of flight is$$t = {2 v \sin(\theta) \over g}$$
$v$ in all equations is the INITIAL velocity of the projectile and $\theta$ is the inclination of the projectile at launch. We can infer that the time of flight is 5 seconds. Therefore, we can use our equation for time of flight to find $\theta$. With $\theta$ being known, we can use the other two equations to find the maximum height and the horizontal distance travelled. Note that $v=32 m/s$ in all cases.

anonymous · Answer

are you sure about the max height formula...it doesn't include time....wouldn't you use another formula then?

anonymous · Answer

From the graph it is clear that at t=2.5sec the particle has velocity which is equal to va that is va is magnitude of horizontal component of the velocity.here at t=2.5sec golf ball is at maximum height..
       hence total time of flight=2*2.5 sec=5 sec...
    (a) as range=horizontal component of the velocity*time of flight=Va*5=16*5=80m
    (b)here UcosX=Va=>cosX=Va/U=16/Vb=16/32=1/2....here U is initial velocity and it is equal to Vb.hence X=60 degree.
                Hmax=(U^2 *sin^2X)/2g.....put the value in this equation and get Hmax....