At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically? Answer in units of m/s.A ball is...

The first ball reaches  the ground after 3 seconds .

Therefore, the time to reach the highest point is 3/2 secs.

The highest height the ball attained = (1/2)g t^2= (1/2)(9.8)(3/2)^2=11.025 m

Let the second ball be thrown with an initial velocity of u m/s with x degree to horizontal

Then its vertical component is usin30.

At the highest point the velocity becomes 0 :

(usinx)-gt=0 . But x=30 degree, given. So,

(usin30)=gt or u(1/2) = gt

Therefore, u= 2gt or t= u/(2g)............(1)

Also the vertical displacement s= (usin30)t-(1/2)t^2

Replacing t from (1) and substituting  s= 11.025m, we get:

11.025= (u/2)(u/(2g) -(1/2)g(u/2g)^2= u^2/(4g)- u^2/(8g)=u^2/(8g)

Therefore, u= sqrt(11.025*8g)= sqrt(11.025*8*9.8)=29.4m/s is the required initial speed of the 2nd ball at 30 degree to horizontal to meet the requirement.