a. h = Yo*t + 4.9g*t^2 = 17.02 m
Yo*2.94 - 4.9*(2.94)^2 = 17.02
2.94Yo - 42.35 = 17.02
2.94Yo = 17.02 + 42.35 = 59.37
Yo = 20.2 m/s = Ver. component of initial velocity.
Dx = Xo*2.94 = 41
Xo = 13.95 m/s = Hor. component of
initial velocity.
tanA = Yo/Xo = 20.2/13.95 = 1.44803
A = 55.4o
Vo = 13.95/cos55.4 = 24.6 m/s[55.4o].
b. Y^2 = Yo^2 + 2g*h
Y^2 = 20.2^2 - 19.6*17.02 = 74.4
Y = 8.63 m/s.
tanA = Y/Xo = 8.63/13.95 = 0.61864
A = 31.74o
V = 13.95/cos31.74 = 16.4m/s[31.74o].
The figure shows the paths of a golf ball your friend drops from the window of her apartment and of the rock you throw from the ground at the same instant. The rock and the ball collide at x = 41.00 m, y = 17.02 m, and t = 2.94 s. If the ball was dropped from a height of h = 59.42 m, determine
a) the initial velocity of the rock,
b) the velocity of the rock at the time of the collision with the ball.
1 answer
1 answer