Calculate Distance a Tiger Will Leap - Tiger Leap Physics

[helpinphysics]

A tiger leaps horizontally from a 12 m high rock with speed of 4.5 m/s. How far from base will she land>

d=vit + 1/2a(t)(t)



12m=4.5m/s*t
t=2.66s

thats where i got stuck

 

[learningphysics]

The 12m is vertical displacement not horizontal.

Split the problem into 2 parts: 1) vertical. 2) horizontal.

The vertical part:

Use the equation: d = v1*t + (1/2)at^2, where v1 is the initial vertical velocity. what is a?

solve for t. Then do the horizontal part...

 

[Doc Al]

Treat horizontal and vertical motion separately. To find the time, use that formula for vertical motion (which is accelerated motion with a = g = 9.8 m/s^2) and solve for t:
$$d = 1/2 a t^2$$

When you find the time, use a formula for the horizontal motion to find the horizontal distance. The horizontal motion is not accelerated.

 

[Astronuc]

The are two parts to this problem.

One is to determine the time to fall 12 m, and the second part is determine who far the tiger travels horizontally during that time.

So part 1 - what is the forumla for ball some distance (height) under the influence of gravity. Use that to find T.

Then the tiger leaps horizontally with speed of 4.5 m/s. How far does the tiger travel in time T with a constant speed of 4.5 m/s.

Please refer to - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

Start with 'Freefall'.

 

[helpinphysics]

i don't think i did this right but is a .00849?

i did 12 m=4.5m/s*2.66s+(1/2)at^2
12=11.97+3.53a
.03/3.53=3.53a/3.53
a=.00849

 

[Astronuc]

Please refer to Doc Al's comment. With respect to learningphysics's post, the initial vertical velocity is zero. One the tiger leaps (horizontally), it starts falling.

One uses the equation of motion in the vertical to find time t or T.

a = g = 9.8 m/s²

Please refer to the hyperphysics page.

 

[helpinphysics]

what do i use to get the horizontal part

 

[helpinphysics]

ok i got that the time is 2.66s

 

[learningphysics]

ok i got that the time is 2.66s

That's not right. Can you show your steps? Did you use the equation Doc Al posted?

 

[Doc Al]

ok i got that the time is 2.66s

How did you arrive at that answer? (Note that that was your original answer, which is still incorrect.)

 

[helpinphysics]

i am really stuck. i don't no what to next

 

[helpinphysics]

i tried to use the equation
but this is how i came up with my answer
d=vit+(1/2)at^2
12=4.5t+(1/2)0t^2
12=4.5m/s*t
t=2.66s

 

[learningphysics]

i am really stuck. i don't no what to next

d = v1*t + (1/2)at^2

v1 = 0, so

d = (1/2)at^2

taking down as positive and up as negative:

-12 = (1/2)(-9.8)t^2

solve for t.

 

[helpinphysics]

but i guess a is 9.81m/s^2

 

[helpinphysics]

t=1.56s

 

[helpinphysics]

so how do i apply that to the horizontal part

 

[helpinphysics]

i found the answer! x=0+4.5*1.56
x=7.02

 

[helpinphysics]

i got the answer
it is 7.02
x=0+4.5*1.56
x=7.02m

 

[learningphysics]

cool. looks right.

 

[helpinphysics]

thank you!

 

[helpinphysics]

i have another quesstion...i did the second part right but I am not sure about the first.

cliff divers push of horizontalll from rock platforms about 35m above the water, but they must clear rocky outcrops at water level that extend out into the water 5m from the base of the cliff directly under their launch point. what minimun speed is necessary to do this? how long are they in the air?

 

[learningphysics]

thank you!

Only one thing. I get 7.04m... this slight difference is due to rounding... don't round till the very end... carry a few extra decimals till the very end when you get the final answer.

 

[helpinphysics]

for how long they are in the air i got 2.7s
d=vit+(1/2)(9.81)t^2
35m=4.905t^2
7.13=t^2
t=2.67 rounded to 2.7s

i can't find the speed

 

[learningphysics]

i have another quesstion...i did the second part right but I am not sure about the first.

cliff divers push of horizontalll from rock platforms about 35m above the water, but they must clear rocky outcrops at water level that extend out into the water 5m from the base of the cliff directly under their launch point. what minimun speed is necessary to do this? how long are they in the air?

once you have the time, just use:

horizontal displacement = horizontal velocity * time

 

[helpinphysics]

i think i found the answer to the first part
d=vit
5m=vi(2.7)
5/2.7=vi(2.7/2.7)
1.851=vi

 

[helpinphysics]

a ball is thrown horizontally from the roof a building 50m tall and lands 45m away from the base. what was the ball's initial speed?

 

[learningphysics]

a ball is thrown horizontally from the roof a building 50m tall and lands 45m away from the base. what was the ball's initial speed?

this one is just like the cliff divers problem...