Question

Jerry the mouse is running along a straight desert road at a constant velocity of 18 m/s. If a certain Tom cat wants to capture Jerry using a net dropped from an bridge that is 15m high, how much time before Jerry is under the bridge, should Tom drop the net? s How far away from the bridge is Jerry when Tom drops the net? m

Answer 1

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:

$y_{f} = y_{0} + v_{oy}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$: is the final height = 0

y₀: is the initial height = 15 m

g: is the gravity = 9.81 m/s²

$v_{0y}$: is the initial vertical velocity of the net = 0 (it is dropped from rest)

$0 = 15m - \frac{1}{2}9.81 m/s^{2}*t^{2}$

$t = \sqrt{\frac{2*15 m}{9.81 m/s^{2}}} = 1.75 s$

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:

$v = \frac{x}{t}$

$x = v*t = 18 m/s*1.75 m = 31.5 m$

Then, Jerry is at 31.5 meters from the bridge when Jerry drops the net.

I hope it helps you!