For a freely falling object dropped from rest, what is the instantaneous speed at the end of the

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2.First, acceleration due to gravity on earth is 9.8 m/s^2 (32.2 f/s^2) so: 

1. v = ta 
v = 5(9.8) 
v = 49 m/s 

v = 6(9.8) 
v = 58.8 m/s 

3. No matter what specific time you have, the acceleration will always be 9.8 m/s^2 due to earths gravity (unless acted upon by an outside force). So the answer to all the questions is 9.8 m/s^2.

4.a.velocity start to decrease.When ball start to fall velocity increases. 
c.acceleration will be equal to gravity(9.8m/s^2) 
e.remain same.

5.You don't seem to have enough information because both the time for the trip down and the time for the trip up depend on the thing you're trying to find out. So you're going to have to derive a pair of simultaneous equations and solve them. 

The equation for a falling object is s = ut + 0.5at^2. The initial velocity is 0, and the acceleration is that due to gravity; call it 10 m/s^2. So s = 5 t^2. 

The journey back is time = speed * distance, so t' = 340 s. That's a different t to the first equation, which is why I've given it a tick mark. I'd sooner give the t in the first equation a subscript 1 and the one in the second a subscript 2, but I don't know how to do that here! 

Anyway, you know t + t' = 3.4, and you should be able to solve those equations!

7.Working formula is 

Vf - Vo = gT 


Vf = velocity of baseball at its peak = 0 
Vo = initial velocity = 25 m/sec. 
g = acceleration due to gravity = 9.8 m/sec^2 (constant) 
T = time for ball to reach its peak 

Substituting values, 

0 - 25 = (-9.8)T 

NOTE the negative sign attached to the acceleration due to gravity. It simply means that the baseball is slowing down as it is going up. 

Solving for T, 

T = 25/9.8 = 2.55 sec. 

And for the baseball to return to the point where it was hit, 

Time = 2.55 * 2 = 5.1 sec. 

<< How high would it move above its starting point? >> 

Working formula, 

Vo = sqrt(2gH) 


H = height at which the baseball will peak 

and all the rest ot the terms have been previously defined 

Substituting values, 

25 = sqrt (2*9.8*H) 

H = 25*25/(2*9.8) 

H = 31.89 m 

<< How fast would it be moving when it returned to its starting point? >> 

Same as its initial velocity of 25 m/sec. 

Please let me know if you nea case for torture by michale levined any clarification. I'm always happy to answer your questions.

qhvatchat (540)
Boston College

Really useful study material!

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