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 79.0 km/h , to his enemy's car, which is going 114 km/h . The enemy's car is 15.1 m in front of the hero's when he lets go of the grenade.

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.

B)Find the magnitude of the velocity relative to the earth.

Explanation / Answer

A)
We can take the car of hero as our frame of reference.
The enemy's car is moving with a velocity, u = 114 - 79 = 35 km/hr = 9.72 m/s.
Consider that after time t, the grenade will reach the enemy's car.
Distance traveled by enemy's car in t time = ut = 9.722 t.
Total distance between the two cars after time t = 15.1 + 9.722 t ...(1)

Horizontal range of the grenade if it is thrown with a velocity v, making an angle 45 is
R = v2 sin (2x45)] / g = v2/g ...(2)
The time the grenade was in the air = [2vsin(45)] / g = 1.414 v/g ...(3)
This is the same time taken the enemy car to cover the distance of ut

Substituting (2) and (3) in (1),
v2/g = 15.1 + 1.414v/g
multiplying with g and rearranging,
v2 - 1.414 v - 147.98 = 0
v = 12.89 m/s.

B)
Vertical component of velocity will be same irrespective of the frame of reference
vy = vsin(45) = 9.11 m/s
Horizontal component will have an additional speed of the car if we are observing from earth.
vx = velocity of car + v (cos(45))
= 79 km/hr + 12.89 x cos(45)
21.94 m/s + 9.11 m/s
= 31.05 m/s
Net velocity = sqrt[vx2 + vy2]
= sqrt[(9.11)2 + (31.05)2] = 32.36 m/s

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