Question

A movie 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

Answer 1

This is a projectile motion problem, so, we use the formula for trajectory:

y =xtanα + gx^2/2v^2(cosα)^2

where 
y is the vertical distance (y = 50 m)
x is the horizontal distance (x=90 m)
α is the angle of trajectory; since it levels of HORIZONTALLY, α = 0°
v is the initial velocity 
g is the acceleration due to gravity which is 9.81 m/s^2

Substituting to the formula,

50 =90tan(0°) + (9.81)(90)^2/2v^2(cos0°)^2
v = 28.2 m/s 

Answer 2

Answer:

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