Question

A skateboarder rolls off a horizontal ledge that is 1.32 m high, and lands 1.88 m from the base of the ledge. What was his initial velocity? (Unit = m/s)

Answer 1

3.62 m/s

Step-by-step explanation:

h = 1.32 m

d = 1.88 m

Let u be the initial horizontal velocity and the time taken by the board to reach the ground is t.

Use second equation of motion in vertical direction

`s=ut+0.5at^(2)`

1.32 = 0 + 0.5 x 9.8 x t^2

t = 0.52 second

The horizontal distance = horizontal velocity x time

1.88 = u x 0.52

u = 3.62 m/s

Thus, the nitial speed of the skate board is 3.62 m/s.

Answer 2

initial velocity is 3.62 m/s

Step-by-step explanation:

given data

high = 1.32 m

length = 1.88 m

to find out

initial velocity

solution

we consider here top height point a and b point at ground and c point at distance 1.88 m away from b on ground

and x is horizontal component and y is vertical component

so at point A initial velocity is Va = Vx i

and at point c velocity = Vc = Vx i + Vy j

first we calculate time taken when it come down by distance formula

distance = 1/2 ×gt² ..............1

1.32 = 1/2 ×(9.8)t²

t = 0.519 sec

so velocity x = distance / time

velocity x =
`(1.88)/(0.519)`

velocity x = 3.622 m/s

so initial velocity is 3.62 m/s