Question

In the latest Indian Jones film, Indy is supposed to throw a grenade from his car, which is going 92.0 km/h , to his enemy's car, which is going 120 km/h . The enemy's car is 14.1 m in front of the Indy's when he lets go of the grenade.

1.If Indy 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.

2.Find the magnitude of the velocity relative to the earth.

Answer 1

Relative velocity = 120 - 92.0 = 28km/hr = 7.777 m/sec

t = time between throwing grenade and landing in enemy car

range of grenade = V x cos 45deg x 2V x sin 45deg / g

= 2Vsquared x cos 45deg x sin 45deg / g

= Vsquared / g (for 45 degrees)

distance = 14.1 + (7.777 x t)

t = 2V x sin 45deg / g

distance = 14.1+ (7.777 x 2V x sin 45deg / g)

Vsquared / g = 14.1 + (11 V /g) (g = 9.81)

V squared = 138.18 + 11 V

V = 138.18 / V + 11

V = 20.43 m/sec (which is 45.7 mph)