if spiderman runs and jumps horizontally from the top of a 200m high building, what does he need his velocity to be to land on a building 100m high, 10m away from him?

v = 2.22 m/s

Explanation:

First we apply the second equation of motion to the vertical motion of the body:

s = Vi t + (1/2)gt²

where,

s = y = vertical distance covered = 200 m - 100 m = 100 m

Vi = V₀y = Vertical Component of Initial Velocity = 0 m/s  (because spider man jumps horizontally, thus his velocity has no vertical component initially)

t = Time Taken to Land on 100 m high building = ?

g = 9.8 m/s²

Therefore,

100 m = (0 m/s)t + (0.5)(9.8 m/s²)t²

t² = (100 m)/(4.9 m/s²)

t = √(20.4 s²)

t = 4.5 s

Now, we analyze the horizontal motion. Neglecting air friction, the horizontal motion is uniform with uniform velocity. Therefore,

s = vt

where,

s = x = horizontal distance covered = 10 m

v = V₀ₓ = Horizontal Component of Initial Velocity = Initial Velocity = ?

Therefore,

10 m = v(4.5 s)

v = 10 m/4.5 s

v = 2.22 m/s

the answer would be 32*f

no sign of dia of fuel cell