In an action-adventure film, the hero is supposed to throw a grenade from his ca

Question

In an action-adventure film, the hero is supposed to throw a grenade from his car, which is going 94.0km/h , to his enemy's car, which is going 102km/h . The enemy's car is 14.9m in front of the hero's when he lets go of the grenade.

Part A

If the hero throws the grenade so its initial velocity relative to him is at an angle of 45 ? above the horizontal, what should the magnitude of the initial velocity be? The cars are both traveling in the same direction on a level road. You can ignore air resistance.

Part B

Find the magnitude of the velocity relative to the earth.

Explanation / Answer

part A

the range is R = 14.9 m

the angle is A = 45o

let u be the initial velocity of the grenade

we know that

R = (u^2 x sin(2A)/g)

or u^2 = (R x g/sin(2A))

or u = (R x g/sin(2A))^1/2

where g = 9.8 m/s^2

part B

the magnitude of the velocity relative to the earth is

v = v1 - u

where v1 = 94.0 km/h = 94.0 x (1000/3600) m/s = 94.0 x (5/18) m/s

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