see
http://www.jiskha.com/display.cgi?id=1423541938
and make the necessary adjustments
http://www.jiskha.com/display.cgi?id=1423541938
and make the necessary adjustments
We can use the equation of motion for the vertical direction:
v(t) = vā + at
where:
v(t) is the velocity of the ball at time t,
vā is the initial velocity,
a is the acceleration,
t is the time.
Initially, the ball is thrown upward, so the initial velocity vā is positive 56 feet per second. The acceleration due to gravity, a, is negative 32 feet per second squared since it acts downward.
Therefore, we have:
v(t) = 56 - 32t
At the highest point of the ball's trajectory, its velocity will be zero. So we can set v(t) = 0 and solve for t:
0 = 56 - 32t
Rearranging the equation, we get:
32t = 56
Dividing both sides by 32:
t = 56/32
Simplifying the fraction, we have:
t = 7/4 = 1.75 seconds
Hence, the ball will be going upwards for 1.75 seconds.