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?

 

[Pyrrhus]

Draw a picture, something that will help you.

 

[umkat]

i did that already...and it didnt help

 

[umkat]

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

 

[Pyrrhus]

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 =)

 

[Pyrrhus]

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??

 

[Pyrrhus]

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

 

[Pyrrhus]

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]

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

 

[Pyrrhus]

did you finish it, umkat?