Amovie stunt driver on a motorcycle speeds horizontally off a 50.0 m high cliff. how fast must the motorcycle leave the cliff-top if it is to land on level ground below, 90.0 m from the base of the cliff where the cameras are

The Horizontal component of initial velocity (V0x) must be 28.18 m/s.

Explanation:

Vertical distance = y = 50m    

Horizontal distance = x = 90 m  

We know that,

x = x0 + V0xt  

where,

x0 = Initial distance in horizontal direction = 0  

V0x = x/t = 90/t          ……… (i)

As,  

y = y0 + V0y + ½ gt^2

where,

y0 = Initial distance in vertical direction = 0

V0y = Vertical component of initial velocity = 0

So,

t = √(50)(2)/(9.8)

t = 3.194 s

Put t = 3.194 s in equation (i),

V0x = 90/3.194

V0x = 28.18 m/s

490 n

f=125n a=2.5m/s2