Question

A gazelle leaps from a cliff 2.5 m high with a speed of 5.6m/s.

a. How much later does
the gazelle hit the ground?

b. How far from the cliff will it land?

c. What is the velocity of the gazelle just before it hits the ground?

Show work, use x and y value chart & use Kinematic Equations

Answer 1

Initial speed of Gazelle is along x direction and its value will be

`v_x = 5.6 m/s`

also its initial height is given as

`y = 2.5 m`

Part a)

now from kinematics along Y direction

`\Delta y = v_y t + (1)/(2) at^2`

as we know that

`\Delta y = 0`

`v_y = 0`

`a = 9.8 m/s^2`

`2.5 = 0 + (1)/(2) (9.8) t^2`

`t = 0.714 s`

Part b)

distance moved horizontally

`\Delta x = v_x t`

as we know that

`v_x = 5.6 m/s`

now we will have

`v_x = 5.6 (0.714) = 4m`

so it will lend at distance of 4 m.

Part c)

final velocity in vertical direction

`v_(fy) = v_y + at`

`v_(fy) = 0 + (9.8)(0.714) = 7 m/s`

`v_x = 5.6 m/s`

so net speed will be

`v^2 = v_x^2 + v_y^2`

`v^2 = 7^2 + 5.6^2`

`v = 8.96 m/s`