PLEASE HELP QUICKLY! 20 points! I will mark best answer as Brainliest!
A pitcher throws a baseball straight into the air with a velocity of 72 feet/sec. If acceleration due to gravity is -32 ft/sec^2, how many seconds after it leaves the pitcher's hand will it take the ball to reach its highest point? Assume the position at time t = 0 is 0 feet.

Respuesta :

t= 2.25 secs

Step-by-step explanation:

Step 1 :

Equation for motion with uniform acceleration is v = u+ at

where v is the final velocity

          u is the initial velocity

          a is the acceleration due to gravity

          and t is the time

Step 2 :

Here , v = 0 because at the highest point final velocity is 0.

u = 72 feet/sec

a = -32 ft/sec^2

We need to find the time t.

Substituting in the equation we have,

0 = 72 -32 * t

=> 32 t = 72

=> t = 72/32 = 2.25 secs