A boy stands 39.2m from a buliding and throws a ball which just passes through a window 19.6m above the ground.Calculate the velocity of projection of the ball..

Question

mayankdevnani · Answer

kinematics......oh!!

anonymous · Answer

The "velocity of projection" part isn't really clear for me. What's that mean?

anonymous · Answer

|dw:1347640639438:dw|$$v_x=vsin \theta$$$$v_y=vcos \theta$$

anonymous · Answer

$$s_x=vsin (\theta) t$$
$$s_y=-\frac{1}{2}gt^2+vcos (\theta) t$$

anonymous · Answer

For$$s_x(39.2)$$Find the time, T.

anonymous · Answer

Then insert the value for T, and you have a system of 2 equations +2 unknowns. Substitute something to solve it.

anonymous · Answer

Use$$v=\frac{s_x}{\sin(\theta)t}$$ and insert it into the second equation (or course, at T s_x=39.2 and s_y=19.6, so put those values in)