Question

A tiger leaps horizontally from a 5 m high rock with a speed of 5 m/s

a) How far from the base of the rock will the tiger land?

b) How much time does it take for the tiger to land?

c) What is the tigers speed when it hits the ground?

Answer 1

Part B.

Given that

Initial Velocity = V0 = 5 m/sec in horizontal direction

V0x = 5 m/sec

V0y = 0 m/sec

Now using 2nd kinematic equation in vertical direction

h = V0y*t + 0.5*ay*t^2

ay = -g = -9.81 m/sec^2

h = -5 m (in downward direction)

So,

-5 = 0*t - 0.5*9.81*t^2

t = sqrt (5*2/9.81) = 1.01 sec

Part A.

Now Range in projectile motion is given by:

Range = V0x*t

R = 5*1.01 = 5.05 m

Part C.

Since there is no acceleration in horizontal direction, So horizontal speed will remain constant

V1x = V0x = 5 m/sec

In vertical direction using 1st kinematic law:

V1y = V0y + ay*t

V1y = 0 - 9.81*1.01

V1y = -9.9 m/sec

Net speed will be

V1 = sqrt (V1x^2 + V1y^2)

V1 = sqrt (5^2 + (-9.9)^2)

V1 = 11.1 m/sec

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