A. h = 0.5g*t^2 = 0.809 m.
4.9t^2 = 0.809
t^2 = 0.1651
t = 0.4063 s.
Xo * t = 18.3 m.
Xo * 0.4063 = 18.3
Xo = 45 m/s. = Velocity of the pitch.
B. h = 0.5g*t^2 = 5.4 m
4.9*t^2 = 5.4
t^2 = 1.1
Tf = = 1.05 = Fall time of each vessel.
Ds = Xo * Tf = 15m/s * 1.05s. = 15.75m.
= Dist. of the spy.
Do = 24m/s * 1.05s. = 25.2 m. = Dist. of official.
Do - Ds = 25.2 - 15.75 = 9.45 m. Apart.
A)...If this pitch were thrown horizontally, the ball would fall .809m by the time it reached home plate, 18.3m away. The acceleration of gravity is 9.81 m/s^2
How fast was the pitch
Answer in m/a
B)... A spy and an official reach the edge of a 5.4m waterfall. The spy's speed is 15 m/s and the officials' speed is 24 m/s.
How far apart will the two vessels be when they land below the waterfall?
The acceleration of gravity is 9.8 m/s^2
Answer in units of m
1 answer