A man takes a 25 ft ladder and props it up against the side of the house so that

Question

A man takes a 25 ft ladder and props it up against the side of the house so that the base of the ladder is 7 ft from the house, and climbs to the top. The ladder starts to slip, and the rate the base is moving away from the house is a constant 3 ft/sec.
1.Find an expression that relates the mans height in feet s above the ground in terms of the distance in feet x that the base of the ladder is from the house.
2.How high above the ground is the man before he starts to fall? How high is he when the ladder is 15 ft from the house?
3Find the mans velocity, ds/dt, in terms of x alone.
4.How fast is Dad falling when the base of the ladder is 15 ft from the house? When the base of the ladder is 24 ft from the house?
5.How far from the house would the base of the ladder be when the man was traveling at the speed of sound(755mi/hr)? (An answer of 25 is not correct)
6. If we assume the top of the ladder is always against the house, what will the mans velocity approach as the base of the ladder approaches 25 ft from the house?
Gravity

Explanation / Answer

Could you please it again with 2 or at max 3 questions in a thread so that its easier to help you explain. Hope you understand.

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