Answer:
v= 20.8 m/s
Explanation:
- Assuming no other forces acting on the ball, from the instant that is thrown vertically downward, it's only accelerated by gravity, in this same direction, with a constant value of -9.8 m/s2 (assuming the ground level as the zero reference level and the upward direction as positive).
- In order to find the final speed 2.00 s after being thrown, we can apply the definition of acceleration, rearranging terms, as follows:
[tex]v_{f} = v_{o} + a*t = v_{o} + g*t (1)[/tex]
- We have the value of t, but since the ball was thrown, this means that it had an initial non-zero velocity v₀.
- Due to we know the value of the vertical displacement also, we can use the following kinematic equation in order to find the initial velocity v₀:
[tex]\Delta y = v_{o} *t + \frac{1}{2} * a* t^{2} (2)[/tex]
- where Δy = yf - y₀ = 15.4 m - 37.4 m = -22 m (3)
- Replacing by the values of Δy, a and t, we can solve for v₀ as follows:
[tex]v_{o} = \frac{(\Delta y- \frac{1}{2} *a*t^{2})}{t} = \frac{-22m+19.6m}{2.00s} = -1.2 m/s (4)[/tex]
- Replacing (4) , and the values of g and t in (1) we can find the value that we are looking for, vf:
[tex]v_{f} = v_{o} + g*t = -1.2 m/s - (9.8m/s2*2.00s) = -20.8 m/s (5)[/tex]
- Therefore, the speed of the ball (the magnitude of the velocity) as it passes the top of the window is 20.8 m/s.
