In an action-adventure film, the hero is supposed to throw a grenade from his ca

Question

In an action-adventure film, the hero is supposed to throw a grenade from his car, which is going 83.0 km/h, to his enemy's car, which is going 122 km/h. The enemy's car is 16.0 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. You can ignore air resistance. Find the magnitude of the velocity relative to the earth.

Explanation / Answer

Vh = 83 kmph = 83*(5/18) = 23.1 m/s

Ve = 122*(5/18) = 34 m/s

along horizantal

velocity relative to earth = vo*cos45 + 83

after t time

for hero car

x1 = ((vo*cos45) + 23.1 )*t......(1)

for enemy car

x2 = 34*t.......(2)

x1 = x2 + 0.016

34*t + 16 = ((v0*cos45) + 23.1)*t.......(3)

along vertical

after t time the grenade falling to ground

y = 0

y = voy*t + 0.5*ay*t^2

0 = (v0*sin45*t) - (0.5*9.8*t^2) ..(4)

solving 3 & 4

t = 3.23 s

v0 = 22.41 m/s = 22.41*(18/5) = 80.7 km/h<-----answer

part(B)

speed relative to earth

vx = vo*cos45 + 83 = 80.7*cos45 + 83 = 140.06 km/h

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