Projectile Motion - Initial velocity

[airi]

1. The problem statement
A child runs down a hill with an angle of 20 degrees and then jumps up at a 20 degree angle above the horizontal. He lands 2.2m down the hill as measured along the hill. What is the child's initial velocity?

The Attempt at a Solution

Okay, I'm not even sure if I have my initial measurements right. This is a projectile problem right? With initial velocity unknown and a 40 degree angle (measured against the ground).

So, first I tried to find the time needed using Vxf=Vx0 + at.
-V0sin40 = Vosin40 - 9.8t
t= 2Vosin40/9.8

Then, i plug that into the Dy=Vy*t
2.2=(Vcos40)(2Vosin40 / 9.8)
10.78 = v^2 * cos40sin40
v=4.68m/s

I tried that answer, but it isn;t correct. What am i doing wrong?

 

[willem2]

This is not the same problem as a projectile problem with a 40 degree launch angle.
The horizontal speed is V_0 cos(20) and not V_0 cos (40) for example.

you need to find equations for the path of the child and for the slope and combine them
to find out where he lands

Find x, and y as a function of t for the child. (use x(t) = x_0 + v_0 t + (1/2) a t^2)

find y as a function of x for the child.

find y as a function of x for the slope.

The child will land if these functions have the same value.
you get an equation with V_0 and x in it, which you can solve for V_0

 

[tomcue]

I would like to first clarify the problem statement. It seems that the child is initially running down a hill at a 20 degree angle, then jumps up at a 20 degree angle above the horizontal. However, the distance of 2.2m down the hill is not clear as to whether it is measured along the hill or straight down. It would also be helpful to know the mass of the child and the duration of the jump.

Assuming that the 2.2m distance is measured straight down, we can use the equations of projectile motion to solve for the initial velocity. First, we need to find the time the child spends in the air, which can be calculated using the equation d=V0t + 1/2at^2, where d is the distance (2.2m), V0 is the initial velocity, a is the acceleration due to gravity (9.8m/s^2), and t is the time. Solving for t, we get t = 0.66 seconds.

Next, we can use the vertical component of the initial velocity (V0y) to find the initial velocity. Using the equation Vfy = V0y - gt, where Vfy is the final vertical velocity (0m/s), V0y is the initial vertical velocity, g is the acceleration due to gravity (9.8m/s^2), and t is the time (0.66 seconds), we can solve for V0y. V0y = 6.47m/s.

Finally, we can use the horizontal component of the initial velocity (V0x) to find the initial velocity. Using the equation Vfx = V0x + at, where Vfx is the final horizontal velocity (V0x), V0x is the initial horizontal velocity, a is the acceleration (0m/s^2), and t is the time (0.66 seconds), we can solve for V0x. V0x = 4.68m/s.

Therefore, the initial velocity of the child is 4.68m/s at an angle of 20 degrees above the horizontal. However, it is important to note that this solution is based on the assumptions made about the problem statement. If there are any discrepancies or missing information, the solution may vary. It is always important to thoroughly understand the problem and its parameters before attempting to solve it.