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. (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?

http://assets.openstudy.com/updates/attachments/4f61de52e4b079c5c6311726-jadefalcon-1331814685036-fig04_40.gif

Answer 1

Let the horizontal distance function be, \[X= X_0+V_0t+\frac{1}{2}a_xt^2\]
where \[ V_0 \text{ is the initial velocity and } X_0 \text{ is the initial distance} \]
From the figure, when Vy=0 => Vx=Va=16
Since Vx is constant, V0=16m/s
\[ X= X_0+V_0t+\frac{1}{2}at^2\ = 0 + (16)(5)+ \frac{1}{2}(0)(5)^2 \]
=80m

(b) Let vertical distance equation be
\[Y= Y_0+V_{0y}t+\frac{1}{2}a_gt^2\]
Maximum height attained at t=2.5
V0 =32, V0x=16 =>
\[ V_{oy}= \sqrt{{32}^2-{16}^2}=16 \sqrt{3} \]
Then at t=2.5
\[Y= Y_0+V_{0y}t+\frac{1}{2}a_gt^2 = 0 + (16)\sqrt{3}(2.5)+ \frac{1}{2}(-9.8)(2.5)^2 \]