Question

In the shot put, a heavy lead weight—the "shot"—is given an initial velocity, starting from an initial elevation approximately equal to the shot putter's height, say, 2.00 m. If v0 = 7.05 m/s, find the horizontal distance traveled by the shot for the following initial angles above the horizontal.

theta 0 =?

theta 40.0 = ?

theta 45.0 = ?

Answer 1

Our equations of motion are now-

and

we solve for (t) in the case where the (y) position of the projectile is at zero (since this is how we defined our starting height to begin with)

Again by applying the quadratic formula we find two solutions for the time. After several steps of algebraic manipulation

The square root must be a positive number, and since the velocity and the cosine of the launch angle can also be assumed to be positive, the solution with the greater time will occur when the positive of the plus or minus sign is used. Thus, the solution is

Solving for the range once again

Substituting -

v=7.05m/s

y0=2m

a) when = 0

Horizontal disrtance travelled= v/g[ 2gy₀]^1/2=4.5m

b) when theta=40,range= 6.756m

c)When theta=45,range=6.593m