Question

A football receiver running straight downfield at 5.50 m/s is 10.0 m in front of the
quarterback when a pass is thrown downfield at 25.0 degree above the horizon. If the
receiver never changes speed and the ball is caught at the same height from which it
was thrown:

(a) Draw a diagram of the motion. Include initial and final positions of the runner.

(b) Find the initial speed of the football.

(c) Find the amount of time the football spends in the air

(d) The distance between the quarterback and the receiver when the catch is made.

Answer 1

a)

b) The motion of football is that of a projectile.

Time of flight of projectile(from O to B) t=2sin/g.................eq(1)

=initial velocity=?

=angle of projection=

g=9.8m/s2

Same time t will be taken for receiver to reach from A to B

time taken for receiver to reach B

t=distance travelled by receiver/velocity of him

Range(OB) of projectile R=sin(2)/g.................(2)

Distance OB in the case of receiver OB=10+vt (v=velocity of receiver=5.5ms)

OB=10+5.5t

t=2sin/g (from equation (1)

OB=10+5.5x(2sin/g).................(3)

Equating equation (2) and (3)

sin(2)/g=10+5.5x(2sin/g)

sin(2x25)/9.8=10+5.5x(2sin25/9.8)

sin(50)/9.8=10+5.5x[2sin(25)/9.8]

0.078=10+0.474

0.078-0.474-10=0

=-(-0.474)±{}/(2x0.078)={14.76m/s0.474+}/0.156

=14.76m/s=initial velocity of football

c)Time of flight of projectile(from O to B) t=2sin/g=(2x14.76xsin25)/9.8=1.273sec

d)Distance OB in the case of receiver OB=10+vt =10+(14.76x1.273)=28.789m