A quarterback throws a pass at an angle of 35° above the horizontal with an initial speed of 25 m/s. The ball is caught by the receiver 2.55 seconds later. Determine the distance the ball was thrown.
A)13 m
B)18 m
C)36 m
D)52 m
E)72 m

Question

Quizplus · Accepted Answer

First, we need to find the horizontal and vertical components of the velocity:

vx = vcosθ = 25cos35° ≈ 20.5 m/s
vy = vsinθ = 25sin35° ≈ 14.4 m/s

Then, we can use the vertical motion equation to find the time it takes for the ball to reach its highest point:

vy = voy - gt
0 = 14.4 - 9.8t
t = 1.47 s

Since the total time in the air is 2.55 s, the time the ball spends falling from its highest point to the receiver's hands is:

t_fall = 2.55 s - 1.47 s = 1.08 s

During this time, the ball will fall a distance equal to:

Δy = voyt_fall - (1/2)gt_fall^2
Δy = 14.4(1.08) - (1/2)(9.8)(1.08)^2
Δy ≈ 6.4 m

Therefore, the receiver must have been 6.4 m below the quarterback when he caught the ball. To find the total distance the ball traveled, we can use the horizontal component of the velocity:

d = vx*t_total
d = 20.5*2.55
d ≈ 52 m

So the answer is D, 52 m.

A Quarterback Throws a Pass at an Angle of 35°