If a baseball leaves the bat at an angle of 45degrees 1m above the ground and clears a wall 3m high 150m away from home plate, what initial velocity would the ball have had if air resistance is negligible
$\displaystyle v_x = vcos(\theta ) = constant;$
$\displaystyle v_y = vsin(\theta )t-gt^2/2+1$
$\displaystyle t=d/v_x$
$\displaystyle v=\sqrt{\frac{\frac{gd^2}{2cos^2(\theta)}}{d\tan\t heta -2}}$
Plug & Chug for 45 degree takeoff angle
$\displaystyle v=\sqrt{\frac{gd^2}{d-2}}$
v = 38.6 m/sec
Yes, I have a typo.... should have wrote $\displaystyle y=vsin(\theta)t-gt^2/2+1$ rather than $\displaystyle v_y = .....$. However, I did not make this error when I actually worked the problem on paper. My expression for v, i.e.
$\displaystyle v=\sqrt{\frac{gd^2}{d-2}}$
is correct