Answer:
a) V_o,y = 0.5*g*t_c
b) V_o,x = D/t_c - v_r
c) V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)
d) Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )
Explanation:
Given:
- The velocity of quarterback before the throw = v_r
- The initial distance of receiver = r
- The final distance of receiver = D
- The time taken to catch the throw = t_c
- x(0) = y(0) = 0
Find:
a) Find V_o,y, the vertical component of the velocity of the ball when the quarterback releases it. Express V_o,y in terms of t_c and g.
b) Find V_o,x, the initial horizontal component of velocity of the ball. Express your answer for V_o,x in terms of D, t_c, and v_r.
c) Find the speed V_o with which the quarterback must throw the ball.
Answer in terms of D, t_c, v_r, and g.
d) Assuming that the quarterback throws the ball with speed V_o, find the angle Q above the horizontal at which he should throw it.
Solution:
- The vertical component of velocity V_o,y can be calculated using second kinematics equation of motion:
y = y(0) + V_o,y*t_c - 0.5*g*t_c^2
0 = 0 + V_o,y*t_c - 0.5*g*t_c^2
V_o,y = 0.5*g*t_c
- The horizontal component of velocity V_o,x witch which velocity is thrown can be calculated using second kinematics equation of motion:
- We know that V_i, x = V_o,x + v_r. Hence,
x = x(0) + V_i,x*t_c
D = 0 + V_i,x*t_c
V_o,x + v_r = D/t_c
V_o,x = D/t_c - v_r
- The speed with which the ball was thrown can be evaluated by finding the resultant of V_o,x and V_o,y components of velocity as follows:
V_o = sqrt ( V_o,x^2 + V_o,y^2)
V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)
- The angle with which it should be thrown can be evaluated by trigonometric relation:
tan(Q) = ( V_o,y / V_o,x )
tan(Q) = ( (0.5*g*t_c)/ (D/t_c - v_r) )
Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )
