In the latest Indian Jones film, Indy is supposed to throw a grenade from his ca

Question

In the latest Indian Jones film, Indy is supposed to throw a grenade from his car, which is going 82.0 km/h , to his enemy's car, which is going 115 km/h . The enemy's car is 14.6 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.

Hint: The grenade moves in projectile motion, and convert the two velocities given in the problem from km/hr to m/s. Being an excellent student of P2, Indy knows that horizontal range of the grenade must equal the distance that the enemies car is ahead at the time the grenade is thrown plus the distance the enemies car travels while the grenade is in the air. This distance is given by v_rel*t, where v_rel the relative velocity of the enemies car relative to the Indy's and t is the time in the air.

Solve for time that the grenade is in the air in terms of v_0.

Use the range equation to get the grenade distance as a function of v_0

Set R=(initial separation)+v_rel*t and get a 2nd order polynomial for v_0. Use quadratic equation to get v_0. This is the magnitude of the velocity vector relative to Indy.

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 granade.

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

Explanation / Answer

A) Let VG be the initial velocity of the grenade relative to the Indy's car. The grenade will move in projectile motion. The equation of the position of the grenade at any time is given by

xG(t) = VG cos(theta)*t,

yG(t) = VG sint(theta)*t - 1/2*g*(t)^2 where theta is the angle of the lanuch. and g is acclearation due to gravity.

To hit the enemy's car the position of the grenade at the time of hit (to) must be equal to the enemy's car position.

Taking the car levels to be the reference in the y direction. i .e ycar = 0 m.

So at time to , yG(to) = 0 i.e VG sint(theta)*to - 1/2*g*(to)^2 = 0 solving this for to

to = 2VGsin(theta)/g

The x position of the enemy's car relative to Indy's car is given by xE(t) = 14.6 + VE*t , where VE is the velocity of the enemy's car in Indy's frame of reference.

Now using xG(t) = xE(t) to obtain the velocity VG of the grenade

14.6 + VE*t = VG cos(theta)*t putting t = to = 2VGsin(theta)/g

VGcos(theta)*2VGsin(theta)/g - VE(2VGsin(theta)/g) - 14.6 = 0

or VGcos(theta)*2VGsin(theta) - VE(2VGsin(theta)) - 14.6*g = 0

g = 9.80 m/s2 , theta = 45 and VE = 115 - 82 = 33 km/h, now we need to convert his into m/s as 14.6 and g is in m/s

VE = 9.167 m/s

therefore (VG)^2 - 12.96VG - 143.08 = 0, solving for VG = 20.08 m/s (other value is negative ignoring that)

Therefore the initial velocity of the grenade relative to the Indy's car = 20.08 m/s or VG = 72.2 km/h

B) The components of velocity of the grenade relative to the ground is given by

Vx = VGx + VI and Vy = VGy where VGx and  VGy are the x and y components of velocity of the grenade relative to the Indy's car and Indy's car is moving at VI = 82.0 km/h

Therefore V = sqrt((Vx)^2 +  (Vy)^2)

V = sqrt((72.2cos(45) + 82)^2 + (72.2sin(45))^2) = 142.51km/h = 143km/h

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.