n an action-adventure film, the hero is supposed to throw a grenade from his car
ID: 2210845 • Letter: N
Question
n an action-adventure film, the hero is supposed to throw a grenade from his car, which is going 97.0km/h , to his enemy's car, which is going 126km/h . The enemy's car is 16.0m in front of the hero's when he lets go of the grenade.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. Answer in km/h
Find the magnitude of the velocity relative to the earth. Answer in km/h
Explanation / Answer
scythian reasoning is perfect but there was one mistake. The relative velocity between the cars is 20 Km/hr or 5.5 meter/sec. The quadratic equation to be solved is: 15.8 + 5.5(2v/g) = v(2v/g) or v^2 - 5.5v - 77.42 = 0 This gives you v = 11.97 m/s. This figure is the horizontal component of velocity. The velocity in the 45 degrees direction is 11.97 m/s divided by Cosine of 45 degrees or 16.92 meters/second measured by the person on car 1. The horizontal velocity of the grenade measured by an external observer is (20 m/s + 11.97 m/2) or 31.97 m/s. The vertical component of the velocity measured by the external observer is 11.97 m/s. So, the total velocity of the grenade in the 45 degrees direction measured by an external observer is SQR (31.97^2 + 11.97^2) or 34.14 m/s. The grenade will hit the enemy car 2.44 seconds after it is thrown. Flight time is 2 x 11.97 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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