In an action-adventure film, the hero is supposed to throw a grenade from his ca
ID: 1587094 • Letter: I
Question
In an action-adventure film, the hero is supposed to throw a grenade from his car, which is going 93.0 km/h , to his enemy's car, which is going 103 km/h . The enemy's car is 15.6 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? Give answer in km/hr.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 (in km/hr)
Show all steps.
Explanation / Answer
Let's make this problem simple by letting v = 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 15.6 meters apart, but the relative velocity of the enemy's car is 2.77 meters per second, so that in time t going at velocity v, the grenade has to travel the horizontal distance of 15.6 + 2.77t. = vt. But we know that t = 2v/g, so that results in a quadratic equation to solve:
15.6 + 2.77(2v/g) = v(2v/g)
or
v^2 - 2.77v - 76.44 = 0
This gives you v = 10.23 m/s. This figure is the horizontal component of velocity. The velocity in the 45 degrees direction is 10.23 m/s divided by Cosine of 45 degrees or 14.46 meters/second measured by the person on car 1. The horizontal velocity of the grenade measured by an external observer is (10m/s + 10.23 m/2) or 20.23 m/s. The vertical component of the velocity measured by the external observer is 10.23 m/s. So, the total velocity of the grenade in the 45 degrees direction measured by an external observer is SQR (20.23^2 + 10.23^2) or 22.66 m/s.
The grenade will hit the enemy car 2.44 seconds after it is thrown.
Flight time is 2 x 10.23 m/s divided by 9.8 m/s^2
The way to prove this value is by substituting the flight time and velocity values in the equation for the distance traveled by both cars.
Distance traveled by car 1 in 2.44 seconds is 74.5 meters
Distance traveled by car 2 is 2.44 seconds is 90.2 meters.
The difference is 90.2m – 74.5m = 15.7 meters, that initially separated both cars.
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