Answer:
1.a) 3.93 m/s
b) 0.80 s
2. a) 8.49 m
b) 0.39 s
Explanation:
1. a) The speed at which the fox leaves the snow can be found as follows:
[tex] v_{f}^{2} = v_{0}^{2} - 2gH [/tex]
Where:
g: is the gravity = 9.80 m/s²
[tex]v_{f}[/tex]: is the final speed = 0 (at the maximum height)
[tex]v_{0}[/tex]: is the initial speed =?
H: is the maximum height = 79 cm = 0.79 m
[tex] v_{0} = \sqrt{2gH} = \sqrt{2*9.80 m/s^{2}*0.79 m} = 3.93 m/s [/tex]
Hence, the speed at which the fox leaves the snow is 3.93 m/s.
b) The time at which the fox reaches the maximum height is given by:
[tex]v_{f} = v_{0} - gt[/tex]
[tex]t = \frac{v_{0} - v_{f}}{g} = \frac{3.93 m/s}{9.80 m/s^{2}} = 0.40 s[/tex]
Now, the time of flight is:
[tex] t_{v} = 2t = 2*0.40 s = 0.80 s [/tex]
2. a) The maximum height the ball reaches is:
[tex] H = \frac{v_{0}^{2} - v_{f}^{2}}{2g} = \frac{(12.9 m/s)^{2}}{2*9.80 m/s^{2}} = 8.49 m [/tex]
Then, the maximum height is 8.49 m.
b) The time at which the ball passes through half the maximum height is:
[tex]y_{f} = y_{0} + v_{0}t - \frac{1}{2}gt^{2}[/tex]
Taking y₀ = 0 and [tex]y_{f}[/tex] = 8.49/2 = 4.245 m we have:
[tex] 4.245 m -12.9m/s*t + \frac{1}{2}*9.80m/s^{2}*t^{2} = 0 [/tex]
By solving the above quadratic equation we have:
t = 0.39 s
Therefore, the time at which the ball passes through half the maximum height when the ball is going up is 0.39 s.
I hope it helps you!
