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Mary spots Bill approaching the dorm at a constant rate of 2.00 m/s on the walkway that passes directly beneath Mary's window, 17.0 m above the ground. When Bill is L = 130 m away from the point below Mary's window she decides to drop an apple down to him. Ignore the effects of air resistance in your calculations.

Required:
How long should Mary wait to drop the apple if Bill is to catch it 1.65 m above the ground?

Answer :

Mickymike92go

Answer:

Mary will have to wait for 63.2 seconds

Explanation:

Time required for the apple to drop from a height of 17.0 m above the ground to 1.65 m above the ground is given by the formula below:

t = √2h/g  where h is height through which the object falls, g is acceleration due to gravity

h = 17.0 - 1.65 = 15.35 m

g = 9.8 m/s²

t = √(2 * 15.35/9.8)

t = 1.77 s or approximately 1.8 s

Time taken for bill to get to the point below Mary's window is given below;

time taken = distance/velocity

distance = 130 m; velocity = 2.0 m/s

time taken by Bill = 130/2.0 = 65 s

Therefore, Mary will have to wait for (65 - 1.8) s = 63.2 seconds

Mary should have to wait for 63.2 seconds

Calculation:

t = √2h/g

here h is height via which the object falls,

g represent acceleration due to gravity

So,

h = 17.0 - 1.65

= 15.35 m

And,

g = 9.8 m/s²

Now

t = √(2 * 15.35/9.8)

t = 1.77 s or approximately 1.8 s

Now the time taken should be

= 130/2.0

= 65 s

So,

= (65 - 1.8) s

= 63.2 seconds

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