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 thrown straight upward and returns to the throwers hand after 3 s in the air. A second ball is thrown at an angle of 30 degrees with the horizontal. The acceleration of gravity is 9.8 m/s^2.
The height of the ball is represented by the equation d = 1/2at^2
The veritcal component of the velocity of the ball from it's maximum hieght to the ground is represented by the equation
vf^2 = vi^2 + 2ad.
The velocity of the ball is repsented by Vr sin 30 = vf
So now we make three substitutions (both d's are the same, max height)
t = 1.5 seconds (time to the max hieght of ball a)
so we get...
vf = ((2a(1/2at^2))^.5)/sin 30
This gives a velocity of 29.4 m/s.
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.