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felipeee4609
27.11.2020 •
Mathematics
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
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Ответ:
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.
Ответ: