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?

Question

I'm not sure how to approach this question.
Thanks.

Answer: 18.6 m/s

Answer 1

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.

Answer 2

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.