Question

In the latest Indian Jones film, Indy is supposed to throw a grenade from his car, which is going 72.0 km/h , to his enemy's car, which is going 128 km/h . The enemy's car is 14.3 m in front of the Indy's when he lets go of the grenade.

Part A

If Indy 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.

Hint: Use Galileo's equation where Frame A is ground, Frame B is Indy's car and the object of interest P is the grenade. Use your answer in part A for speed and given direction to calculate the x and y components of the velocity with respect to Indy's car. The velocity of Indy's car with respect to the ground is straightforward to write in x and y components.

Answer 1

Part A:

The range of projectile at an angle of 45^∘ above the horizontal with an initial speed u will be

$R=\frac{u^{2}\sin 2\times 45^{o} }{g}=\frac{u^{2} }{g}$

If you consider that Indi´s car is at rest and then enemy's car will go at a speed of (128 - 72) or 56 Km/h i.e. 15.56 m/s. The vertical component of speed of the grenade is u sin 45^o or $\inline u/\sqrt{2}$ . Therefore, the time of flight will be

$T=\frac{2u\sin 45^{o} }{g}=\frac{\sqrt{2}u}{g}$

Therefore, the distance traveled by the enemy´s car will be 15.56 T which must equal to the range of the projectile.

$\frac{\sqrt{2}u}{g}\times 15.56=\frac{u^{2}}{g}$

$\therefore u=15.56\sqrt{2}=22\ m/s=79.2\ Km/h$

The magnitude of the initial velocity will be 79.2 Km/h.

Part B:

The velocity of the grenade along the horizontal and vertical direction (as the angle is 45^o) are 15.56 m/s or 56 Km/h. The velocity of Indy´s car in horizontal direction is 72 Km/h. Therefore, the grenade velocity with respect to earth along horizontal direction will be (72 Km/h + 56 Km/h) or 128 Km/h. The vertical component of grenade velocity with respect to Earth will be 56 Km/h. Therefore, the resultant velocity with respect to earth will be

$V=\sqrt{128^{2}+56^{2}}=139.7\ Km/h=38.8\ m/s$

the magnitude of the velocity relative to the earth will be 139.7 Km/h