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 91.0 km/h , to his enemy's car, which is going 125 km/h . The enemy's car is 15.4 m 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

solution:

Let's make this problem simple by letting v =VcosQ vertical and horizontal velocities, since the grenade is being thrown at a 45° angle.

Then the time it takes for the grenade to rise and fall back to the level of the cars is 2v/g, where g is 9.81 m/s². This is because the time for the grenade's velocity v to drop to 0 is just v - gt = 0, and then it has to get back down again, right?

Now, the cars are 15.4 meters apart, but the relative velocity of the enemy's car is 20km/h(5.55m/s) so that in time t going at velocity v, the grenade has to travel the horizontal distance of 15.8 + 5.55t. = vt.

But we know that t = 2v/g, so that results in a quadratic equation to solve:

15.8 + 5.55(2v/g) = v(2v/g) here g =9.8 m/s^2

v^2- 5.55v-77.42 =0 solve for v =12m/s and earlier we assumed that horizontal velocity v =VcosQ here Q =45

so V =16.97 m/s ans A

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