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) )
).
(
). Even without knowing the precise value of the pure water resistivity, we can estimate the ratio between the pure water resistivity and the silver resistivity by comparing the two orders of magnitude:
