A BALL IS THROWN STRIAGHT UPWARD AND RETURMS TO THE THROWER'S HAND AFTER 3 SECONDS IN THE AIR. A SECOND BALL IS THROWN AT AN ANGLE OF 30 DEGREES WITH THE HORIZONTAL. AT WHAT SPEED MUST THE 2ND BALL BE THROWN SO THAT IT REACHES THE SAME HIEGHT AS THE ONE THORWN VERTICALLY?
I wish I could help but I am taking Biology this year!Sorry but um....check your Physics book!
Also,thanks for the math help!
First, find the vertical speed of ball 1.
hf=hi + vi*t +1/2 g t^2
0=0 + vi*3 + 1/2 (-9.8)*9
solve for Vi.
Then, on ball 2
ThatVi= VoSin30
solve for Vo.
To solve this problem, we need to find the initial velocity (Vo) of the second ball.
The first ball reaches the same height as the initial position after 3 seconds. We can find the vertical speed of the first ball using the equation:
hf = hi + vi*t + (1/2)g*t^2
where hf is the final height (which is equal to the initial height, hi), vi is the initial velocity (which we want to find), t is the time (which is 3 seconds), and g is the acceleration due to gravity (-9.8 m/s^2).
Substituting the known values into the equation:
0 = 0 + vi*3 + (1/2)(-9.8)(3^2)
Simplifying the equation:
0 = 3vi - 44.1
Rearranging the equation to solve for vi:
vi = 44.1/3
Therefore, the vertical speed of the first ball is approximately 14.7 m/s.
Now, let's consider the second ball. The vertical component of its initial velocity (Vo) can be found using the equation for the vertical component of velocity:
Vi = Vo * sin(θ)
where Vi is the vertical component of velocity (which we found to be 14.7 m/s from the first ball), and θ is the angle at which the second ball is thrown (which is 30 degrees).
Substituting the known values into the equation:
14.7 = Vo * sin(30)
Simplifying the equation:
14.7 = (Vo * √3) / 2
Multiplying both sides by 2/√3:
Vo = 2 * 14.7 / √3
Therefore, the second ball must be thrown at a speed of approximately 17.0 m/s.