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
