Baseball player A bunts the ball by hitting it in such a way that it acquires an

Question

Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.1 m/s parallel to the ground. Upon contact with the bat, the ball is 1.4 m above the ground. Payer B wishes to duplicate this bunt, in so far as he also wants to give the ball a velocity parallel to the ground and have his ball travel the same horizontal distance as player A's ball does. However, player B hits the ball when it is 1.8 m above the ground. What is the magnitude of the initial velocity that player B's ball must be given?
m/s

Explanation / Answer

time of f;ight for A :

h =gt^2 /2

t =sqrt(2h/g) = sqrt(2 x1.4 /9.8) = 0.535 sec

d = vt 0.535 x 1.1 = 0.588 m

for B.

t = sqrt(2 x 1.8 / 9.8) = 0.606 sec

v = d/t = 0.588 / 0.606 = 0.970 m/s

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.