Answer to Question #257785 in Quantum Mechanics for lulu

Question

           Natalya throws a shot put. The shot leaves her hand 1.9 m above the ground at a velocity of 8.0 m s-1 at an angle of 37o to the horizontal. Calculate the time between the shot leaving her hand and hitting the ground. Start by working out the initial vertical component of the shot put motion.

Answer 1

The time upward:

$$t=v_y/g=v\sin\theta/g=0.49\text{ s}$$

The height above the point where the ball was thrown:

$$h=gt^2/2=1.18\text{ m}.$$

The total max. height above the ground:

$$H=h+h_0=1.18+1.9=3.08\text{ m}.$$

The time it took to fall from height H:

$$\tau=\sqrt{2H/g}=0.79\text{ s}$$

The total time:

$$T=t+\tau=1.28\text{ s}.$$