A) If the hero throws the grenade so its initial velocity relative to him is at

Question

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.

B) Find the magnitude of the velocity relative to the earth.

In an action-adventure film, the hero is supposed to throw a grenade from his car, which is going 95.0 km/h , to his enemy's car, which is going 109 km/h. The enemy's car is 16.4 m in front of the hero's when he lets go of the grenade.

Explanation / Answer

Given that

The car going with a speed is (vh) =95km/h =26.389m/s

The enemey car which is going with a speed(ve) =109km/h =30.278m/s

Now we consider to 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 the acceleration due to gravity 9.81 m/s².

Now the car which is 16.4m apart but but the relative velocity of the enemy's car is 14km/h =3.889m/s so that in time t going at velocity v, the grenade has to travel the horizontal distance of 16.4 + 3.889t = vt

But we know that t = 2v/g, so that results in a quadratic equation becomes is

16.4 + 3.889(2v/g) = v(2v/g)

now the aboveequation becomes 2v2-7.778v-160.884=0

Solving the above quadratic equation we get

v =11.12s

b)

Now horizontal velocity of the grenade measured by an external observer is (26.389 m/s + 11.12 m/2) or 37.509 m/s. The vertical component of the velocity measured by the external observer is 11.12 m/s. So, the total velocity of the grenade in the 45 degrees direction measured by an external observer is

v =SQR (37.509^2 + 11.12^2) =39.122m/s.

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