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...
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.