Question

As a quarterback throws the football, it leaves his hand at a height of h and at an initial speed of v0 at an angle θ above the horizontal. The receiver can’t get to the football in time and it falls onto the field. What is the ball’s speed as it hits the ground? Solve this problem in two ways: (a) using kinematics, explicitly calculating the trajectory the ball takes (b) using conservation of (kinetic + potential) energy

Answer 1

consider the motion of football in Y-direction

v_oy = initial velocity in Y-direction = v_o Sinθ

a_y = acceleration = - g

Y = vertical displacement = - h

v_fy = final velocity as it hits the ground = ?

using the equation

v²_fy = v²_oy + 2 a_y Y

v²_fy = (v_o Sinθ)² + 2 (- g) (- h)

v²_fy = (v_o Sinθ)² + 2 gh eq-1

consider the moton in horzontal direction or x-direction

v_ox = initial velocity in x-direction = v_o Cosθ

a_x = acceleration = 0 m/s²

X = horizontal displacement  

v_fx = final velocity as it hits the ground = ?

using the equation

v²_fx = v²_ox + 2 a_x X

v²_fx = (v_o Cosθ)² +  2 (0) X

v²_fx = v²_o Cos²θ eq-2

using pythagorean theorem , net speed as it hits the ground is given as

v_f² = v²_fx + v²_fy

using eq-1 and eq-2

v_f² = v²_o Cos²θ +  (v_o Sinθ)² + 2 gh

v_f² = v²_o (Cos²θ + Sin²θ) + 2 gh

v_f = sqrt(v²_o + 2 gh)

b)

Case : using conservation of energy

at the point of launch of football , initial total energy is given as

initial total energy = kinetic energy + potential energy

initial total energy = mgh + (0.5) m v_o²

at the point where football hits, final total energy is given as

final total energy = kinetic energy

final total energy = (0.5) m v_f²

using conservation of energy

final total energy = initial total energy

(0.5) m v_f² = mgh + (0.5) m v_o²

(0.5) v_f² = gh + (0.5) v_o²

v_f² = 2gh + v_o²

v_f = sqrt(2gh + v_o²)