# Projectile Motion Leap Question

umkat
QUick Question:
A tiger leaps horizontally from a 6.5 m high rock with a speed of 3.5 m/s. How far from the base of the rock will the tiger land?

I started this problem by trying to find the vert. and horz. components, but I was not given an angle to use sin and cos with...how do i start this problem? Homework Helper

umkat
i did that already...and it didnt help

umkat
would the angle be 90 degrees since the tiger is jumping horizontally?

Homework Helper
umkat said:
would the angle be 90 degrees since the tiger is jumping horizontally?

no, it will be 0 degrees, of course I'm assuming max height is his initial position, it seems like that to me.

umkat
thanks.
i found how many seconds it will take (1.15)
but i still don't know how to find out the distance
sorry for being dumb =)

Homework Helper
Well at the beginning he only has a Vx component, and remember it's always constant from the beginning to the end, so why don't you use it to calculate the distance?

umkat
would the final answer be V(initial)*t = 3.5 m/s (1.15s) = 4.025 m...
so 4 meters away??

Homework Helper
If that's the time when it lands, then yes.

umkat
Thanks for your help...I just found this site tonite and signed up and it is real helpful. Thanks some much for the help, I'm sure I will continue to visit it again. Thanks

Homework Helper
I've too found this site recently and yes i believe it is a great resource for students and teachers alike, it's always great to be of help.

umkat
okay I have one more question: Romeo is chunking pebbles gently up to Juliet's window, and he wants the pebbles to hit the window with only a horizontal component velocity. He is standing at the edge of a rose garden 4.5 m below her window and 5.0 m from the base of the wall. How fast are the pebbles going when they hit her window? I worked it out and got 5.6 m/s...is that what the answer is if anyone out there wants to work it out? thanks.

CellCoree
umkat said:
thanks.
i found how many seconds it will take (1.15)
but i still don't know how to find out the distance
sorry for being dumb =)

sorry to go back to a question you already answered, but how did you find the time it will take?

i found that
V(0)x = 3.5m/s
V(0)y = 0 m/s

would you use this to find the time?
x(0) would be 4.5 since it's the initial height right?
x(t) = x(0) + v(0)t + 1/2(a)(t)^2?
0 = 4.5 + 3.5t + 1/2(-9.8)(t)^2

is that how you solved for t?

umkat
i said that D=1/2(9.8)(1.15) which equals 5.63
to find t i did t =square root of 2d/g or 2(5.63)/9.8 which equals 1.15

Homework Helper
did you finish it, umkat?