Question

A golfball is hit from ground level with an initial vertical velocity of 64 ft/s. Use the function h(t) = −16t2 + v0t + h0, where h(t) is the height of a golfball hit with initial vertical velocity v0 from initial height h0 after t seconds in the air, to find the time at which the golfball will hit the ground. after 4 s after 5 s after 3 s after 6 s

Answer 1

Answer:

It is seen that the height of the golf ball is maximum at t= 3 sec that is 50 m above the ground.

The ball hits the ground at t= 4 seconds that is it's distance from the ground is  zero meters.

Step-by-step explanation:

h(t) = −16t2 + v0t + h0

Putting the values

Initial velocity = V0 = 64m/s

time = 4, 5,3,and 6 seconds

initial height = h0 = ground level= 0 m

h(t) = −16t2 + v0t + h0

h(4) = −16(4)2 + 64(4) + 0

h(4)= 0m

h(t) = −16t2 + v0t + h0

h(5) = −16(5)2 + 64(5) + 0

h(5)= - 400 + 320= -80m

h(t) = −16t2 + v0t + h0

h(3) = −16(3)2 + 64(3) + 0

         = -144+ 192= 50m

h(t) = −16t2 + v0t + h0

h(6) = −16(6)2 + 64(6) + 0  

       = -576+ 384= -192m

It is seen that the height of the golf ball is maximum at t= 3 sec that is 50 m above the ground.

The - ( negative) sign at different times shows that the ball may follow a path  lower than the ground level.

The ball hits the ground at t= 4 seconds that is it's distance from the ground is  zero meters.