In the latest Indian Jones film, Indy is supposed to throw a grenade from his car, which is going 43.8 m/s, to his enemy's car, which is going 55.8 m/s. The enemy's car is 15.2 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.
My set up here is
horizontal distance = V * cos45 * Vj * t
[where V is the initial velocity and Vj is Jones' car velocity]
distance between cars = 15.2 + (Vj - Ve) * t
[where Ve is the enemy's car velocity]
I set these two equations equal to each other to find the time it takes the grenade to reach the other car.
(Vcos45)(43.9)(t) = 15.2 + (43.8 - 55.88) * (t)
Then I'm not sure if I'm head in the right direction...
B) Find the magnitude of the velocity relative to the earth.
I'm also not sure how to start this portion.
Thank for you the explanation!
But I'd like to ask, how come the last equation isn't
V^2/g = 15.2 + 12(2Vsin45/g)?
They are the same, since
2 sin 45 = sqrt2
drwls - Your answer isn't correct either. The horizontal distance traveled is V*cos 45*t + Vj*t, and the time it takes for the grenade to reach the ground is SQRT(2Vsin45/g)
Where does V^2/g come from?
A) You are definitely on the right track for part A! You have set up the correct equations to find the time it takes for the grenade to reach the other car. Now, let's solve the equation to find the value of time, and then use that to determine the magnitude of the initial velocity.
Let's simplify the equation you set up:
(Vcos45)(43.9)(t) = 15.2 + (43.8 - 55.8) * (t)
Using trigonometric identity cos45 = sqrt(2)/2, we can rewrite the equation as:
(V(sqrt(2)/2)(43.9)(t) = 15.2 + (-12)(t)
Now, let's isolate the variable t:
(V(sqrt(2)/2)(43.9)(t) + 12(t) = 15.2
Taking common terms out:
[(V(sqrt(2)/2)(43.9) + 12) * t = 15.2
Now, we can solve for t:
t = 15.2 / [(V(sqrt(2)/2)(43.9) + 12)]
Now that we have the value of time, we can substitute it back into one of the original equations to find the magnitude of the initial velocity:
(Vcos(45))(43.8)(t) = 15.2 + (43.8 - 55.8) * (t)
Let's further simplify this equation:
(V(sqrt(2)/2))(43.8)(t) = 15.2 - 12t
(V(sqrt(2)/2))(43.8)(15.2 / [(V(sqrt(2)/2))(43.9) + 12]) = 15.2 - 12[15.2 / [(V(sqrt(2)/2))(43.9) + 12]]
Now, solve for V in this equation. Note that you need to use numerical methods or approximation techniques to solve this equation algebraically.
B) To find the magnitude of the velocity relative to the Earth, we can use the Pythagorean theorem. As the car is moving in the same direction, the velocity of the grenade is the sum of the velocity of the car and the initial velocity of the grenade relative to the car.
Let's say the magnitude of the initial velocity of the grenade is Vg, and the magnitude of the velocity of the car is Vc.
The magnitude of the velocity relative to the Earth (Ve) is given by:
Ve = sqrt((Vc + Vg)^2)
Now, substitute the values of Vc and Vg that you have:
Ve = sqrt((43.8)^2 + V^2)
Simplify this equation further:
Ve = sqrt(1916.44 + V^2)
So the magnitude of the velocity relative to the Earth is given by the square root of (1916.44 + V^2).
Your equations are not correct, but I want to thank you for being one of the very few who show their work here. We wish everyone would do that.
The horizontal distance the grenade travels from the launch point, in earth-based coordinates, is
V*cos45 + Vj*t
The distance between cars is
15.2 + (Ve - Vj)*t = 15.2 + 12.0 t
The way to get t is NOT to set those distances equal.
Consider a coordinate system moving with Jones' car. The grenade moves
V^2/g before hitting the ground when lauched at a 45 degree angle in that coordinate system. The time it takes to hit the ground is
t = 2Vsin45/g.
After that time, the separation of cars is
15.2 + 12 t.
For the grenade to hit the enemy car,
V^2/g = 15.2 + 12*sqrt2*V/g
Solve that quadratic equation for V. Take the posite root.
B) V is relative to the car in which Jones is travelling. Relative to the Earth, add 43.8 m/s to that.