Askateboarder 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)

initial velocity is 3.62 m/s

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 =

velocity x = 3.622 m/s

so initial velocity is 3.62 m/s

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s