Question

A football receiver running straight downfield at 4.60 m/s is 11.5 m in front of the quarterback when a pass is thrown downfield at 31.0° above the horizon.

If the receiver never changes speed and the ball is caught at the same height from which it was thrown, find the football's initial speed, the amount of time the football spends in the air, and the distance between the quarterback and the receiver when the catch is made.

(a)

the football's initial speed (in m/s)

(b)

the amount of time the football spends in the air (in s)

(c)

the distance between the quarterback and the receiver when the catch is made (in m)

Answer 1

a)

consider the motion along the X-direction

V_ox = initial velocity of ball along X-direction = v Cos31

t = time of travel of the ball

v_r = speed of reciever = 4.60 m/s

X = horizontal distance travelled = 11.5 + 4.60 t

along the horizontal direction ,

X = 11.5 + 4.60 t

(v Cos31) t = 11.5 + 4.60 t

t = 11.5/(v Cos31 - 4.60) eq-1

consider the motion along Y-direction

Y = vertical displacement = 0

V_oy = v Sin31

a = acceleration = - 9.8

using the equation

Y = V_oy t + (0.5) a t²

0 = (v Sin31) t + (0.5) (- 9.8) t²

using eq-1

0 = (v Sin31) (11.5/(v Cos31 - 4.60) ) + (0.5) (- 9.8) (11.5/(v Cos31 - 4.60) )²

v = 14.3 m/s

using equation 1

t = 11.5/(v Cos31 - 4.60) = 11.5/((14.3) Cos31 - 4.60)

t = 1.5 sec

c)

X = 11.5 + 4.60 t

X = 11.5 + 4.60 (1.5) = 18.4 m