Question

A quarterback throws a football toward a receiver with an initial speed of 20 m/s at an angle of 30∘ above the horizontal. At that instant the receiver is 20 m from the quarterback. In (a) what direction and (b) with what constant speed should the receiver run in order to catch the football at the level at which it was thrown?

Answer 1

Answer:

a. The runner must run 15.35 m in the direction of the ball. b. 7.52 m/s

Explanation:

This is an example of projectile motion. We first calculate the range of the football to know where it is going to land.

Given,

initial speed of football,u = 20 m/s

Angle of projection, θ = 30°

a. We calculate the range R = u²sin2θ/g = 20²sin(2 × 30)/9.8 = 400sin60/9.8 = 400 × 0.8660/9.8 = 346.41/9.8 = 35.348 m ≅ 35.35 m

Since the receiver starts 20 m from the quarterback, the distance he runs in order to catch the ball is d = (35.35 - 20) m = 15.35 m.

So the runner must run 15.35 m in the direction of the ball.

b. The time it takes him to run the distance d = 15.35 m and catch the ball is the time it takes the projectile to land at the height he catches the ball which is given by t = 2usinθ/g = (2 × 20 × sin30)/9.8 = (40 × 0.5)/9.8 = 20/9.8 = 2.041 s ≅ 2.04 s

To find the constant speed at which the receiver must run in order to catch the ball at the same level of throw, we use s = vt + 1/2at²

since a = 0 (constant speed), s = vt ⇒ v = s/t

s = d = 15.35 m and t = 2.041 s

v = 15.35 m/2.041 s = 7.521 m/s ≅ 7.52 m/s

Answer 2

Answer:

a) In order to catch the ball at the level at which it is thrown in the direction of motion.

b)Speed of the receiver will be 7.52m/s

Explanation:

Calculating range,R= Vo^2Sin2theta/g

R= (20^2×Sin(2×30)/9.8 = 35.35m

Let receiver be(R-20) = 35.35-20= 15.35m

The horizontal component of the ball is:

Vox= Vocostheta= 20× cos30°

Vox= 17.32m/s

Time taken to coverR=35.35m with 17.32m/s will be:

t=R/Vox= 35.35/17.32

t= 2.04seconds

b)Speed required to cover 15.35m at 2.04seconds

Vxreciever= d/t = 15.35/2.04 = 7.52m/s