Thread: finding the initial minimum velocity....

ok so the problem is this...

a projectile is fired from ground level at an angle of 12 degrees with the horizontal. the projectile is to have a range of 200 ft. what is the initial minimum velocity?

i dunno if i did this right, but i got V(o)*cos(12) = 200 ft and solved for V(0)? so V(0) = 204.468 ft/s is this right? can someone check and see if i did this right? it seems too easy....

thanks in advance...

First of all, v0 cos(12)= 200 ft makes no sense at all. The left hand side of the equation is a speed and the right side is a distance- they can't be equal. Assuming no air resistance, V0cos(12) is the horizontal component of initial velocity and remains so: the horizontal distance traveled is V0cos(12)t. Set that equal to 200 feet and solve for t as a function of V0:
V0cos(12)t= 200 so t= 200/(V0cos(12).

Now, look at the vertical motion. The acceleration due to gravity is -g or about -32.2 feet/sec^2. The vertical speed is V0 sin(12)-32.2 t feet/sec and the vertical height, at time t, is -V0 sin(12)- 16.1t^2. Put your expression for t into that, set it equal to 0 (where the projectile hits the ground again and solve for V0.