A tiger leaps horizontally from a 5.5 m high rock with a speed of 3.1 m/s. ?

Question

How far from the base of the rock will shde land in meters?

Answer 1

consider the x and y directions separately.

Y-direction:

initial velocity = 0m/s

acceleration= -9.8m/s^2 (acceleration due to gravity)

displacement= -5.5m (displacement and acceleration are negative because they are in the negative y direction)

time = ?

find t:

s=vit+1/2at^2

(s=displacement, vi = initial velocity, a=acceleration, t=time)

-5.5=0t+1/2(-9.8)t^2

-5.5=-4.9t^2

-5.5/-4.9=t^2

t=1.06 seconds

This will also be the time in the x direction.

X direction:

initial velocity=3.1m/s

acceleration=0

displacement=?

time= 1.06

find s (displacement):

s=vit+1/2at^2

s=3.1*1.06+1/2(0)t^2

s=3.29m

:. Tiger lands 3.29 m away from rock.

Answer 2

The horizontal and vertical parts of the projectile are independent. The time it takes for the tiger to hit the ground is found by the equation;

distance = 1/2 a t^2

distance = 5.5 meters

a= accel of gravity = 9.81 m/s/s

we don't know t

5.5 = 1/2 (9.81) t^2

solve for t and you get t=1.06 seconds

We now know the tiger is in the air 1.06 seconds. The distance from the base is found by the equation;

distance=velocity*time = 3.1 m/s*1.06 seconds =3.3 meters