Thread: angle and velocity

ive been trying this problem for about an hour now, but i cant seem to figure it out.

a cannon shoots a cannonball at an angle of 37 degrees from a cliff 40m high. with a horizontal velocity of 35.7m/s, find the angle the cannonball strikes the ground, and velocity it strikes the ground.

thanks

Originally Posted by johnnyc418

ive been trying this problem for about an hour now, but i cant seem to figure it out.

a cannon shoots a cannonball at an angle of 37 degrees from a cliff 40m high. with a horizontal velocity of 35.7m/s, find the angle the cannonball strikes the ground, and velocity it strikes the ground.

thanks

If the horizontal component of intial velocity is 35.7, then the vertcal component is 35.7 tan(37). You can then calculate that the vertical velocity at each time t seconds after being fired is - 9.8t+ 35.7 tan(37) while the horizontal velocity is constant at 35.7.

The vertical height of the cannonball, taking its initial position as h=0 is $\displaystyle h= -4.9t^2+ 35.7tan(37)t$. The ground at the bottom of the cliff is at h= -40 so you can find the time the cannonball hits the ground by solving $\displaystyle -4.9t^2+ 35.7 tan(37) t= -40$.

You can then calculate the velocity vector by putting that time into v_x= 35.7 and $\displaystyle v_y= -9.8t+ 35.7 tan(37)$.

The angle at which the cannonball hits the ground is given by $\displaystyle tan^{-1}\left(\frac{v_x}{v_y}\right)$.