Answer:
a) t = 1.75 s
b) x = 31.5 m
Explanation:
a) The time at which Tom should drop the net can be found using the following equation:
[tex] y_{f} = y_{0} + v_{oy}t - \frac{1}{2}gt^{2} [/tex]
Where:
[tex] y_{f}[/tex]: is the final height = 0
y₀: is the initial height = 15 m
g: is the gravity = 9.81 m/s²
[tex]v_{0y}[/tex]: is the initial vertical velocity of the net = 0 (it is dropped from rest)
[tex] 0 = 15m - \frac{1}{2}9.81 m/s^{2}*t^{2} [/tex]
[tex]t = \sqrt{\frac{2*15 m}{9.81 m/s^{2}}} = 1.75 s[/tex]
Hence, Tom should drop the net at 1.75 s before Jerry is under the bridge.
b) We can find the distance at which is Jerry when Tom drops the net as follows:
[tex] v = \frac{x}{t} [/tex]
[tex] x = v*t = 18 m/s*1.75 m = 31.5 m [/tex]
Then, Jerry is at 31.5 meters from the bridge when Jerry drops the net.
I hope it helps you!
