In the latest Indian Jones film, Indy is supposed to throw a grenade from his car, which is going 93.0 km/h, to his enemy's car, which is going 116 km/h. The enemy's car is 16.9 m in front of the Indy's when he lets go of the grenade.

If Indy throws the grenade so its initial velocity relative to him is at an angle of 45 degrees 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.

Hint: The grenade moves in projectile motion, and convert the two velocities given in the problem from km/hr to m/s.

Being an excellent student of P2, Indy knows that horizontal range of the grenade must equal the distance that the enemy's car is ahead at the time the grenade is thrown plus the distance the enemy's car travels while the grenade is in the air. This distance is given by vrel∗t,vrel∗t, where vrelvrel the relative velocity of the enemies car relative to the Indy's and t is the time in the air.

Solve for the time that the grenade is in the air in terms of v0.v0.

Use the range equation to get the grenade distance as a function of v0.v0.

Set R=(initial separation) + vrel∗tvrel∗t and get a 2nd order polynomial for v0.v0. Use the quadratic equation to get v0.v0. This is the magnitude of the velocity vector relative to Indy.

Answer rating: 100% (QA)

In projectile motion problem solving it is not unusual that the target is also in motion In these oc... View the full answer