What horizontal speed must a pumpkin be thrown to hit a car 13.4 meters away from a building which stands 10.4 meters tall? A) 1.5 m/s
B) 2.1 m/s
C)6.1 m/s
D) 8.9 m/s

V₀ₓ = 9.2 m/s

Nearest

D) 8.9 m/s

Explanation:

First we find the time taken by the pumpkin to hit the car. For that purpose we apply 2nd equation of motion to the pumpkin:

h = V₀y t + (1/2)gt²

where,

h = height of building = 10.4 m

V₀y = vertical component of initial speed = 0 m/s

t = time = ?

g = 9.8 m/s²

Therefore,

10.4 m = (0 m/s)(t) + (1/2)(9.8 m/s²)t²

t² = (10.4 m)(2)/(9.8 m/s²)

t = √[2.122 s²]

t = 1.45 s

Now, we analyze horizontal motion for horizontal component of initial velocity. We assume air friction to be zero so that the horizontal motion is uniform. Therefore,

s = V₀ₓ t

where,

s = horizontal distance between building and car = 13.4 m

V₀ₓ = Horizontal Component of Initial Velocity = ?

Therefore,

13.4 m = V₀ₓ(1.45 s)

V₀ₓ = 13.4 m/1.45 s

V₀ₓ = 9.2 m/s

answer: b.

explanation:

truck y has a greater velocity than truck x.