The launching speed of a certain projectile is 8.0 times the spped it has at its maximum height. Calculate the elevation angle at launching.

The speed of the projectile can be broken into two velocity components. Vy and Vx. At maximum height Vy is equal to 0. The velocity vector V=(Vy^2+Vx^2)^1/2

We also know, that the initial speed of the projectile is 8 times the amount of max height.

The initial velocity of the projectile is Vo, and that can be broken into two components as well. Vox and Voy. From vectors, Vo=(Vox^2+Voy^2)^1/2

Vo=8 Vx... substitute we get... Vox^2 + Voy^2 = 64*Vx^2.

Tan(x)=Voy/Vox... Voy= Tan(x)*Vox.

Substitute again.

Vox^2 + (Tan(x)*Vox)^2 = 64*Vx^2.

Now one final equation. Vx=Vox because gravity does not act in the X direction but only the y.

So we divide by Vox^2 and we get

1+ (Tan(x))^2=64

The height h, in meters, of a projectile t seconds, after it is fired from a cannon is given by h=kt-1/2gt^2. Suppose we assume that k= 50 and g= 9.8 m/s^2, when is the projectile at a height of 20 meters?

You should have started your own thread marijoi.

use the equation given to you.

h = kt - \frac12 gt^2

Use the values that were given to you, h = 20, k = 50, g = 9.8.

20 = 50t - \frac12 (9.8)t^2

You now get a quadratic. Rearrange, and then solve this quadratic.