Question

A stunt performer is to run and dive off a tall platform and land in
a net in the back of a truck below. Originally the truck is directly
under the platform. It starts forward with a constant acceleration
a at the same instant that the performer leaves the platform. If the
platform is H above the net in the truck, then the horizontal
velocity u that the performer must have as he leaves the platform is​

Answer 1

Answer:

u = a√(H/2g)

Explanation:

Let say Horizontal Speed = U

Vertical Speed = 0

Vertical Distance = H

S = ut + (1/2)at²

Time Taken  

H = 0 + (1/2)gt²

=> t = √(2H/g)

Horizontal Distance = Distance Traveled by Truck

S = ut + (1/2)at² = (1/2)a2H/g = aH/g

Distance Traveled by Performer

= u * √(2H/g)

u * √(2H/g)  = aH/g

=> u = a√(H/2g)

Answer 2

Answer:

U=under root a[H/2g]

Explanation:

Hope you can make the sum by your own mind