Word problem not making any sense. - StupidHomework.com

**Word problem not making any sense.** : 03-17-2008

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.

03-19-2008

I solved it, but I have a another one.

Sam shot his model rocket up into the air and counted the seconds it stayed in flight. He noticed that the rocket seemed to be at the same height as the top of a television antenna at 3 and 6 seconds. Find the initial speed of the model rocket and the height of the television antenna.

www.last.fm/user/shmads/ · 03-19-2008

There should be a formula involving initial height, angle, and initial speed that you need to use for this question. Use this formula, and know that when x=3 and 6, the y-value is equal.

IDK what that formula is though, its more of a physics thing.

03-19-2008

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.

03-19-2008

Quote:

Originally Posted by Git

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.

yeah exactly, there are always two of the same point one traveling upwards and one downwards

03-20-2008

Quote:

Originally Posted by Nugwin

I solved it, but I have a another one.

Sam shot his model rocket up into the air and counted the seconds it stayed in flight. He noticed that the rocket seemed to be at the same height as the top of a television antenna at 3 and 6 seconds. Find the initial speed of the model rocket and the height of the television antenna.

Yeah...it's just a manipulation of the formula that v(final)=v(start)+at. It's actually the derivative of that equation.

As for your problem...you could deduce the fact that the rocket would reach its zenith at 4.5 seconds. So, without going through the problem, you could use the acceleration due to gravity on earth, and the 4.5 seconds that you know, as well as the final velocity being zero because it's at a zenith, to find the starting velocity.