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In an action film hero is supposed to throw a grenade from his car, which is going 90.km/h, to his enemy's car, which is going 110 km/h. The enemy's car is 15.8 m 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(degree) 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. Ignore air resistance.
Find the magnitude of the velocity both relative to the hero and relative to the earth
Dear sachin,
At first we need to convert all the velocities into m/s:
We will describe the motion in the hero's reference frame. In this reference frame the velocity of the enemy's car is
And the grenade is released with the launch angle . We introduce the magnitude of initial velocity of grenade (relative to the hero's car) as v.
After the grenade is released it is moving according to the equations of projectile motion. Along axis x we have motion with constant velocity:
Along axis y we have motion with constant acceleration:
In these equations at the initial moment of time the grenade has zero coordinates.
We can also write down the equation of enemy's car motion:
Now we need to write down the condition that the grenade hits the enemy's car:
At some moment of time the x coordinate of grenade should be equation to the x-coordinate of enemy's car and at the same moment of time the y-coordinate of grenade should be equal 0.
Then we have:
Then we just need to solve the system of two equations with two unknown variables:
Then
And
This is the velocity of the grenade relative to the hero.
We can also find the magnitude of velocity relative to the earth. The x-component of this velocity is
The y-component of the velocity is
Then the magnitude is
This is the magnitude of the velocity of grenade relative to the earth.
Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.
All the best.
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