1) A hawk is carrying a mouse 40m above the ground, flying 10 m/s horizontally toward her nest of babies who are 8m above the ground. When she is 100m horizontally from her nest, the mouse starts struggling. She can sense that in 5 seconds the mouse will jump out of her grasp, so she starts accelerating parallel to the ground toward the nest. What should be her constant acceleration be so that when the mouse slips, it lands in the nest? Om 8m 100m

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Please answer question 1, everything needed is in the picture

1) A hawk is carrying a mouse 40m above the ground, flying 10 m/s horizontally toward her
nest of babies who are 8m above the ground. When she is 100m horizontally from her
nest, the mouse starts struggling. She can sense that in 5 seconds the mouse will jump
out of her grasp, so she starts accelerating parallel to the ground toward the nest. What
should be her constant acceleration be so that when the mouse slips, it lands in the
nest?
40m
8m
100m
Transcribed Image Text:1) A hawk is carrying a mouse 40m above the ground, flying 10 m/s horizontally toward her nest of babies who are 8m above the ground. When she is 100m horizontally from her nest, the mouse starts struggling. She can sense that in 5 seconds the mouse will jump out of her grasp, so she starts accelerating parallel to the ground toward the nest. What should be her constant acceleration be so that when the mouse slips, it lands in the nest? 40m 8m 100m
Expert Solution
Step 1

Here, the hawk is in a horizontal motion, and when the mouse slips it will be in a vertical motion. So we can divide the given data according to vertical and horizontal components.

Since the mouse should fall from 40 m to 8m, it takes time to reach the net. That can be found by using the constant acceleration formula.

y=viyt+12ayt2.....1here y=8-40=-32 mviy=0 ay=-g=-9.8m/s2.From 1,-32=0-0.5×9.8×t2t=324.9=2.55 s

The mouse will take 2.55 s to reach the nest, once it slips from the hawk.

The remaining time left for the hawk is  t'=5-t=5-2.55=2.45 s

 

 

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