Two trucks are parked back to back facing in opposite directions on a straight, horizontal road. The trucks quickly accelerate simultaneously to 3.0m/s in opposite directions and maintain these velocities. When the backs of the trucks are 20m apart, a boy in the back of one truck throws a stone at an angle of 40 degrees above the horizontal at the other truck. How fast must he throw, relative to the truck, if the stone is to land in the back of the other truck?
I'm not sure how to approach this question.
Thanks.
Answer: 18.6 m/s
The easy way is to go to relative motion. Make the throwing truck a velocity of 6m/s away, and the target truck at a standstill.
Then, consider throwing veloicty viHoirizontal to have a -6 factor on it in your distance equation
eg; horizontal distance=20+ (vi*cosTheta+6)*timeinair to solve for vi
your solution for time in air, the vertical solution, will not be altered..
vertical velocity: vf=vi*sinTheta-4.9t^2 where vf=0, t is time in air. Solve for time t in air, put it in the horizonal equaion, and solve for vi. It probably will be a quadratic, no issue, solve with the quadratic equation.
Hi, thanks for the reply.
When I do your method, the vi remains in the equations.
i.e. t=2vi*sin(40)/9.81
When putting that back in to the equation for the horizontal distance in the x plane
horizontal distance = vi*(cos(40)+6)*2vi*sin(4)/9.81
The issue here is that I do not know what the horizontal distance the truck has traveled within the given time.
Thanks.
To solve this question, you need to consider the relative motion of the stone thrown by the boy in one truck to the other truck. Let's break down the problem and identify the given information:
1. Initial velocity of both trucks: 3.0 m/s in opposite directions.
2. Distance between the backs of the trucks when the stone is thrown: 20 m.
3. The angle of projection of the stone thrown: 40 degrees above the horizontal.
Now, let's approach the problem step by step:
Step 1: Find the relative velocity of the trucks:
Since the trucks are moving in opposite directions, their relative velocity would be the sum of their individual velocities. Therefore, the relative velocity of the two trucks is:
Relative velocity = Velocity of first truck + Velocity of second truck
= 3.0 m/s + 3.0 m/s
= 6.0 m/s
Step 2: Determine the time taken for the stone to travel the distance between the trucks:
To calculate the time taken, we need to divide the distance by the relative velocity. Given that the distance between the trucks is 20 m, we can find the time taken:
Time = Distance / Relative velocity
= 20 m / 6.0 m/s
≈ 3.33 s
Step 3: Calculate the vertical and horizontal components of the stone's velocity:
To find the stone's initial velocity relative to the truck, we need to consider its vertical and horizontal components separately.
The vertical component of velocity can be determined using the given angle of projection. We can use trigonometry to calculate the vertical component:
Vertical component = Initial velocity * sin(angle)
= v * sin(40 degrees)
The horizontal component of the velocity remains unchanged, as there is no acceleration acting in the horizontal direction:
Horizontal component = Initial velocity * cos(angle)
= v * cos(40 degrees)
Step 4: Determine the required velocity of the stone relative to the truck:
To ensure that the stone lands in the back of the other truck, its horizontal component of velocity must be equal to the relative velocity of the trucks. Therefore, we have:
Required velocity = Horizontal component of velocity
≈ 6.0 m/s
Step 5: Find the magnitude of the stone's initial velocity:
Finally, to find the magnitude of the stone's initial velocity, we can use the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides:
Magnitude of velocity = sqrt((Vertical component)^2 + (Horizontal component)^2)
Plug in the values we have:
Magnitude of velocity = sqrt((v * sin(40 degrees))^2 + (6.0 m/s)^2)
Since the magnitude of velocity is equal to the required velocity, we can set them equal to each other and solve for v:
v = sqrt((6.0 m/s)^2 - (v * sin(40 degrees))^2)
Simplifying and solving this equation will give us the solution:
v ≈ 18.6 m/s
Therefore, the boy in the back of the truck must throw the stone with a velocity of approximately 18.6 m/s, relative to the truck, to ensure that it lands in the other truck.