Answer:
a) 48.5 ft/s
b) 36.5 ft
c) -80.3 ft/s
Explanation:
a)
The equation of motion of the ball is :
y(t) = -16.1 ft/s^2 * t^2 + Vo*t
Where Vo is the initial velocity
If y(5s) = - 160 ft:
-160 ft = -16.1 ft/s^2 * (5 s)^2 + Vo*(5s)
Solving for Vo
Vo = (16.1*25- 160) ft / 5s = 48.5 ft/s
b)
To answer this question we must first know when the velocity became zero, at this time is when the ball was at its highest point.
v(t) = -32.2 ft/s^2 * t + Vo
t = Vo/32.2ft/s^2 = 1.5 s
And now, the highest point which the ball reached is given by:
y(1.5s) = -16.1 ft/s^2 * (1.5)^2 + Vo*(1.5s)
y(1.5s) = 36.52 ft
c)
We now need the time at which y(t') = -64 ft
-64 = -16.1*t'^2 + 48.5*t'
By means of the quadratic formula, we find that
t' = 4.00498 s ≈ 4 s
And the velocity at t = 4s is:
v(4s) = -32.2 ft/s^2 * 4s +48.5 ft/s = -80.3 ft/s
