A quarterback throws a football with an initial speed of 22m/s at an angle of 63
ID: 1504464 • Letter: A
Question
A quarterback throws a football with an initial speed of 22m/s at an angle of 63 above the horizontal toward a receiver who is at a distance of 30 m from the quarterback. The receiver runs and catches the ball at the same level it was thrown.
a) What are the x and y components?
b)How long is the ball in the air when it gets caught at the same level that it is thrown?
c)How far does the ball travel?
d)In what direction and what constant speed should the reciever run in order to catch the football?
Explanation / Answer
(a)
Vx = 22 * cos(63)
Vx = 9.99 m/s
Vy = 22 * sin(63)
Vy = 19.60 m/s
(b)
time ball is in the air,
S = vy*t + 1/2*gt^2
0 = 19.60 * t - 1/2*9.8*t^2
t = 4 s
(c)
Distance travelled by ball,
Sx = Vx * t
Sx = 9.99 * 4.0 m
Sx = 39.96 m
(d)
Distance = 39.96 - 30.0 = 9.96 m
t = 4 s
Speed, V = 9.96/4
V = 2.49 m/s
Direction = Away from the quarterback !!