Question

a ball is thrown vertically upward with a speed of 8.0 m/s (a) How high does it rise? (b) how long does it take to reach its highest point? (c) How long does it take to hit the ball to take to hit the ground after it reaches its highest point? (d) What is its velocity when it returns to the level from which it started?

Answer 1

Because acceleration is constant we can use our simplified kinematics equations.

For parts a and b:

initial velocity = v(0) = 8m/s
initial height = y(0) = 0m (note choosing this as our starting height doesn't affect the answer)
final velocity = v(t) = 0m/s
final height = y(t) (we need to find this)
acceleration = a = g = -9.81m/s^2 (constant, note this is negative because it acts in the direction opposite to our initial velocity which has been defined as positive)

Using the equation v(t) = v(0) + a*t (this should be in your textbook)

=> t = [v(t) - v(0)]/a
= [0 - 8]/(-9.81)
= 0.816s

this gives us the answer to b. The ball takes 0.816s to reach its highest point.

Using the equation y(t) = y(0) + v(0)*t + 0.5*a*t^2
= 8*t + 0.5*g*t^2
= 8*(0.816) + 0.5*(-9.81)*(0.816)^2 (subbing in t = .816s from part b)
= 6.86m

this gives us the answer to a.

For c and d the approach is very similar we need only to consider that
v(0) = 0m/s and
y(0) = 6.86m
and keep in mind that both velocity and acceleration are now negative.