Question

A spring gun, held horizontally 1.70 m above the ground, fires a 0.010 kg marble so that it lands a horizontal distance of 7.60 m away. If the gun is pointed straight up, and the same marble is fired, how high (in meters) will it rise?

Work shown in next post!

Answer 1

So I started off with the time to fall:
\[h=\frac{ 1 }{ 2 }g t ^{2} = t \sqrt{2h/g}\] and I ended up getting 0.589s

next I found the velocity in the horizontal direction

v= distance / time = 7.60m/0.589 = 12.903 m/s

next I used v as the Vi for the vertical velocity and calculated the time it takes to get to the top:

Vf = Vi-gt

and finally calculated h=Vit - 1/2gt^2

my final answer was 10.52m but according to my homework its wrong. Help please! :(

Answer 2

steps 1 and 2 look good. step 3 looks odd.

for that step, use the 3rd formulation of Newon's laws of motion, one that a lot of people seem to shy away from but that is extremely extremely useful and time-saving for things like this:

\[v^2 = u^2 + 2ax\]

v is zero, u is the 12.903 m/s you calculated, a = -g.

remember also that you might need to add in the initial height above the ground, depends what the question requires.

reality check too. 10.52m in the vertical direction compared to 7.60 m in horizontal just looks wrong.