Question

Casey loves baseball, and one night has a vivid dream about saving his team in the last inning of the game when they are three runs down and with bases loaded. Casey steps to the plate, points the bat upward toward left field, and then proceeds to hit the ball out of the park to achieve the Grand Slam needed to win the game. The crowd goes wild! He later learns that the ball traveled approximately 200 meters before landing in the parking lot. Assuming that the ball was initially hit 1.0 meter above the ground, and at angle 45 degrees above horizontal, what was the initial speed of the ball? Ignore air resistance and assume that the ground is level.

	A.	64 m/s
	B.	44 m/s
	C.	90 m/s
	D.	55 m/s

Answer 1

Range in projectile motion is given by:

R = V0x*t

V0x = Initial horizontal velocity = V0*cos 45 deg

t = R/V0x = 200/(V0*cos 45 deg)

t = 282.84/V0 = time taken by ball to land in parking lot

Now Using 2nd kinematic equation in vertical direction

h = V0y*t + (1/2)*a*t^2

h = vertical displacement = -1.0 m

V0y = Initial vertical velocity = V0*sin 45 deg

a = acceleration due to gravity = -9.8 m/s^2

So,

-1.0 = V0*sin 45 deg*t + (1/2)*(-9.8)*t^2

-1.0 = V0*sin 45 deg*200/(cos 45 deg*V0) - 4.9*(200/cos 45 deg*V0)^2

-1.0 = 200 - 4.9*200^2/(2*V0^2)

9.8*200^2/V0^2 = 201

V0 = sqrt (9.8*200*200/201)

V0 = 44.162 m/s = 44 m/s

Correct option is B.