Question

In the latest Indian Jones film, Indy is supposed to throw a grenade from his car, which is going 81.0 km/h , to his enemy's car, which is going 109 km/h . The enemy's car is 13.4 m in front of the Indy's when he lets go of the grenade.
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.

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

Answer 1

Answer:

  A.  18.21 m/s  or  65.6 km/h

  B.  135.5 km/h

Step-by-step explanation:

A. The vertical speed of the grenade is ...

  vv = v·sin(45°) = v/√2

The time in air is given by the equation for ballistic motion:

  h = -4.9t² +vv·t = 0

  t(vv -4.9t) = 0 . . . . . factor out t

  v/√2 = 4.9t

  v = t·4.9√2 . . . . . . solve for v

__

The horizontal distance the grenade travels in t seconds is ...

  d = vh·t = v·cos(45°)·t = vt/√2 = (t·4.9√2)(t/√2) = 4.9t² . . . . meters

1 m/s = 3.6 km/h, so the horizontal distance the target travels is ...

  x = (109 - 81.0)/3.6·t + 13.4 = 70t/9 +13.4 . . . . meters in t seconds

We want the target distance to be the same as the grenade's distance, so the time required to hit the target is ...

  d = x

  4.9t² = (70/9)t +13.4

  4.9(t² -(100/63)t) = 13.4

  4.9(t -50/63)² = 13.4 +4.9(50/63)² ≈ 16.486

  t = 50/63 + √(16.486/4.9) ≈ 2.62793 . . . seconds

This corresponds to a launch speed of ...

  v = (4.9√2)t ≈ 18.21 . . . . meters/second

__

B. As we said, 1 m/s = 3.6 km/h, so the launch speed is about 65.558 km/h at an angle of 45° relative to the direction of travel. The magnitude of the velocity relative to the earth (ve) will be this vector added to 81.0 km/h in the direction of travel. The (horizontal, vertical) components of that sum are ...

  (81.0, 0) + 65.558(cos(45°), sin(45°)) ≈ (127.357, 46.357) km/h

The magnitude is the Pythagorean sum:

  ve = √(127.357² +46.357²) ≈ 135.5 . . . . km/h