Question

Use a(t) =−32 feet per second squared as the acceleration due to gravity. a ball is thrown vertically upward from the ground with an initial velocity of 56 feet per second. for how many seconds will the ball be going upward?

Answer 1

Since the ball is moving by uniformly accelerated motion, its vertical velocity at time t is given by
$v(t)= v_0 - a t$
where we took upward as positive direction, and where $v_0$ is the initial velocity, a the acceleration and t the time.

The instant at which $v(t)=0$ is the instant when the ball reverses its velocity (from upward to downward). This means that the difference between the time t at which v(t)=0 and the instant t=0 is the total time during which the ball was going upward:
$0=v_0 - at$
By plugging numbers into the equation, we find
$t= \frac{v_0}{a}= \frac{56 ft/s}{32 ft/s^2}=1.75 s$