Projectile Motion with wall

[rsmseys]

Homework Statement

A projectile, initial speed v = 20 m/s is supposed to hit a target D = 28.9 m away. However an H = 10 m high wall is placed halfway between the launching point and the target. What is the appropriate elevation angle? (in degrees).

Homework Equations

http://en.wikipedia.org/wiki/Equations_of_motion#Equations_of_uniformly_accelerated_linear_motion

The Attempt at a Solution

I got the answer of around 44 to 45 degrees. The correct answer is said to be closer to 65 degrees. Any help in clarifying which answer is right would be helpful, along with solutions. Thanks.

 

[The legend]

and how did you reach your answer?

According to PF rules no solns can be given... just help or advice.

 

[rsmseys]

v = at
20sin(x) = 9.8t
20sin(x) / 9.8 = t
s = 1/2(u +v)t
10 = 1/2 * (20sin(x))*(20sin(x))/9.8

Solving for x, the answer for x (the angle) is 44.427 degrees.

 

[zgozvrm]

Not quite, but you're on the right track.

Work it backwards with your result and you'll see that although you just cleared the wall, you missed the target by nearly 12m.

 

[rwisz]

I would first check to make sure that all the given values are accurate and consistent. Then, I would use the equations of motion to solve for the appropriate elevation angle. In this case, we can use the equation y = y0 + v0y * t - 1/2 * g * t^2, where y is the vertical distance, y0 is the initial vertical distance, v0y is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.

First, we can find the time it takes for the projectile to reach the wall by setting y = 10 m and solving for t. This gives us t = 1 second. Then, we can use this time in the equation y = y0 + v0y * t - 1/2 * g * t^2 to solve for the initial vertical velocity, v0y. This gives us v0y = 10 m/s.

Next, we can use the equation x = x0 + v0x * t to solve for the horizontal distance, x0, which is the distance from the launching point to the wall. This gives us x0 = 10 m. Then, we can use this distance and the given distance to the target, D = 28.9 m, to solve for the horizontal velocity, v0x. This gives us v0x = 20.9 m/s.

Finally, we can use the equation tan(theta) = v0y / v0x to solve for the elevation angle, theta. This gives us theta = 28.6 degrees. However, this is the angle at which the projectile would hit the wall. To hit the target, the projectile must travel another 28.9 m after hitting the wall. This means that the total horizontal distance the projectile must travel is 28.9 m + 10 m = 38.9 m. Using this new distance in the equation v0x = x / t, we can solve for the new horizontal velocity, v0x. This gives us v0x = 31.4 m/s.

Finally, we can use this new horizontal velocity and the equation tan(theta) = v0y / v0x to solve for the elevation angle, theta. This gives us theta = 64.5 degrees, which is closer to the given correct answer of 65 degrees.

In conclusion, the