1. 1.( 2 points) In diving to a depth of 995 m, an elephant seal also moves 839

Question

1.

1.( 2 points) In diving to a depth of 995 m, an elephant seal also moves 839 m due east of his starting point. What is the magnitude of the seal’s displacement?

2.( 4 points) A spacecraft is traveling with a velocity of v0x = 6870 m/s along the +x direction. Two engines are turned on for a time of 772 s. One engine gives the spacecraft an acceleration in the +x direction of ax = 1.93 m/s2, while the other gives it an acceleration in the +y direction of ay= 7.46 m/s2. At the end of the firing, what is a) vx and b) vy?

3.( 4 points) A puck is moving on an air hockey table. Relative to an x, y coordinate system at time t = 0 s, the x components of the puck’s initial velocity and acceleration are v0x = +7.8 m/s and ax = +7.1 m/s2. The y components of the puck’s initial velocity and acceleration are v0y = +9.4 m/s and ay = -4.3 m/s2. Find (a) the magnitude v and (b) the direction of the puck’s velocity at a time of t = 0.50 s. Specify the direction.

4.( 4 points) A rocket is fired at a speed of 92.0 m/s from ground level, at an angle of 70.0o above the horizontal. The rocket is fired toward an 55.2-m high wall, which is located 40.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?

5.( 4 points) Stones are thrown horizontally with the same velocity from the tops of two different buildings. One stone lands four times as far from the base of the building from which it was thrown as does the other stone. Find the ratio of the height of the taller building to the height of the shorter building.

Explanation / Answer

a) dispalcement = sqroot (995^2 + 839^2) = 1301.5168 m apprx

b)

Vx= 6870 + 772 (1.93) = 8359.96 m/s

Vy = 7.46 (772)= 5759.12 m/s

c) Vx= 7.8 +7.1 (0.5)= 11.35 m/s

Vy= 9.4 + (-4.3) (0.5) = 7.25 m/s

v = sqroot (11.35^2 + 7.25^2) = 13.47 m/s apprx

angle = tan^-1 ( 7.25/ 11.35)= 32.569 degree apprx

4) Uy = 92 sin 70 = 86.45 m/s apprx

Ux= 92 cos 70

time taken to reach 40 m = 40/ 92 cos 70 = 40 / (31.465 ) = 1.27 sec

h= 86.45 ( 1.27) - 0.5 (9.8) ( 1.27)^2= 109.793- 7.903= 102 m apprx

height cleared = 46.8 m apprx

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