A baseball is hit at a height of 1 m with an initial velocity v, at 37° above the horizontal. It just clears a barrier of height
20 m at a horizontal distance of 32 m. Find (a) vo ; (b) the time to reach the barrier.
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Explanation:
The ball travels distance D in ∆t = 2.60 s. Therefore, the x component of the initial velocity is D =v
ox
⇒ ∆t = D/∆t = 15.4 m/s Due to the symmetry of projectile motions, it takes 1.2 seconds the ball to reach the initial height after passing the top of the wall downward. Therefore, the total flight time is t
tot
= 1.2 + 2.6 +1.2 = 5 s The horizontal range is R = v
ox
t
tot
= 77.0 m
By symmetry, the ball reaches the peak of the motion at time t
top
= t
tot
/2 = 2.5 s At the peak, the y component of the velocity is zero. Thus we have 0 = v
oy
– gt
top
⇒ v
oy
= gt
top
= 24.5 m/s The height of the wall is equal to the height of the ball at t = 1.2 s h = y (t = 1.2 s) = v
oy
+v
oy
t – (1/2)gt
2
= 23.3 m
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