Quote:
Originally Posted by Nugwin  A tennis ball is shot vertically upward from a launcher with an initial velocity of 64 ft/s. When will the ball return to the ground? How far off the ground will the ball be after 1 second? After 3 seconds? (Hint: acceleration of gravity is 32 ft/s).
Ok, so I think the formula I should use is
s=(v1)t-1/2gt^2
Distance above starting point(s) is equal to initial velocity(v1) multiplied by time elapsed(t) minus one half the product of accerlation of gravity(g) and time elapsed(t) squared
According to the problem:
s=what im solving for
v1=64
t=1 and 3
g=32 ft/s
I plug these numbers in and I get...
s=64(1)-1/2(32)(1^2)
s=64(3)-1/2(32)(3^2)
Simplified...
s=64-16
s=48
s=192-144
s=48
Ok, we got that settled. But what the hell...logically...a ball can't be at the same height after 1 second and after 3 seconds. Also how do I figure out how to find when the ball will return to the ground? Do I keep on plugging in numbers until s=0? I'm not sure here. Help guys.. Thanks so much. |
It is at the same height. After 1 second, it is 48 feet in the air heading up, and after 3 seconds it is 48 off the ground, but traveling downwards.