Question

A football quarterback is traveling straight backward at a speed of 5.1 m/s when he throws a pass to a receiver 15.5 m straight downfield.

(a) If the ball is thrown at an angle of 25° relative to the ground and is caught at the same height as it is released, what is its initial speed relative to the ground in m/s?

(b) How long does it take to get to the receiver in s?

(c) What is its maximum height above its point of release in m?

Could you explain the answer and show detailed work? Thanks! :)

Answer 1

a)

v_o = initial speed of ball relative to ground

Consider the motion of the ball along the horizontal direction

v_ox = velocity in horizontal direction = v_o Cos25

a_x = acceleration = 0 m/s²

t = time taken

X = horizontal displacement = 15.5 m

Using the kinematics equation

X = v_ox t + (0.5) a_x t²

15.5 = ( v_o Cos25) t = + (0.5) (0) t²

t = 15.5 /( v_o Cos25)                                                           Eq-1

Consider the motion of the ball along vertical direction

v_oy = velocity in vertical direction = v_o Sin25

a_y = acceleration = - 9.8 m/s²

t = time taken

Y = vertical displacement = 0 m

Using the kinematics equation

Y = v_oy t + (0.5) a_y t²

Using eq-1

0 = ( v_o Sin25) ( 15.5 /( v_o Cos25)) + (0.5) (- 9.8) (15.5 /( v_o Cos25))²

v_o = 14.08 m/s

b)

Using eq-1

t = 15.5 /( v_o Cos25)   

t = 15.5 /( 14.08 Cos25)   

t = 1.2 sec

c)

Consider the motion from initial point of launch to highest point

y = vertical displacement = maximum height = h_max

a_y = acceleration due to gravity = - 9.8 m/s²

v_fy = final velocity at highest point = 0 m/s

v_oy = initial velocity = v_o Sin25

Using the equation

v_fy² = v_oy² + 2 a_y y

0² = (v_o Sin25)² + 2 (- 9.8) h_max

0² = (14.08 Sin25)² + 2 (- 9.8) h_max

h_max = 1.81 m