1. A constant force of 393 N acts on a spacecraft of mass 7850 kg that has an in

Question

1. A constant force of 393 N acts on a spacecraft of mass 7850 kg that has an initial velocity of 31 m/s. How far has the spacecraft traveled when it reaches a velocity of 4870 m/s?

2. A bullet is fired upward with a speed v0 from the edge of a cliff of height h (see figure below). Ignore air drag. Assume the bullet is fired straight up in (a) and (b) and straight down in (c).

(a) What is the speed of the bullet when it passes by the cliff on its way down? (Use the following as necessary: v0.)

(b) What is the speed of the bullet just before it strikes the ground? (Use the following as necessary: v0, g, and h.)

(c) If the bullet is instead fired downward with the same initial speed v0, what is its speed just before it strikes the ground? (Use the following as necessary: v0, g, and h.)

Question Part Points Submissions Used A constant force of 393 N acts on a spacecraft of mass 7850 kg that has an initial velocity of 31 m/s. How far has the spacecraft traveled when it reaches a velocity of 4870 m/s? A bullet is fired upward with a speed v0 from the edge of a cliff of height h (see figure below). Ignore air drag. Assume the bullet is fired straight up in (a) and (b) and straight down in (c). What is the speed of the bullet when it passes by the cliff on its way down? (Use the following as necessary: v0.) What is the speed of the bullet just before it strikes the ground? (Use the following as necessary: v0, g, and h.) If the bullet is instead fired downward with the same initial speed v0, what is its speed just before it strikes the ground? (Use the following as necessary: v0, g, and h.) A rock is dropped from a tree of height 26 m into a lake (depth 4.4 m) below. After entering the water, the rock floats gently down through the water at a constant speed of 1.3 m/s to the bottom of the lake. What is the total elapsed time?

Explanation / Answer

1)a=F/m=393/7850=0.05m/s^2

u=31m/s; v=4870 m/s

v^2=u^2+2as

s=(v^2-u^2)/2a=2.37*10^8 m

2)a)change in potential energy when it passes by the cliff on its way down=0

so change is kinetic energy=0, so speed will remain same=v0.

b)let speed=v,writing equation between points when it passes by the cliff on its way down and when it has reachd ground.

v^2=v0^2+2gh

v=(v0^2+2gh)^0.5

c)when bullet is fired downwards its velocity when it reaches ground=v'

v'2=v^2=v0^2+2gh

v'=(v0^2+2gh)^0.5.

3)time to fall from tree=t

26=0.5*g*t^2

t=2.30s.

time taken in water=4.4/1.3=3.38s.

total time=2.3+3.38=5.68s

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