I'm having problems understanding 5 of my homework problems and was wondering if anyone could help me with them.(adsbygoogle = window.adsbygoogle || []).push({});

1) Two forces are applied to an old tree stump, to pull it out of the ground. One rope pulls due east with a force of 2100 N. A second force is 45 degrees South of East with a magnitude of 2800 N. Calculate the magnitude and direction of the resultant force.

My solution: c^2 = a^2 + b^2

sqrt 2100^2 + 2800^2 = 3500 N

theta = tan^-1 (2800/2100)= 53.13 degrees

2) A roofer working 12.0 meters above the ground tosses a piece of scrap wood off the roof. Just for fun he tosses it nearly straight up with a velocity of 4.3 m/s.

a) Calculate the velocity when it hits the ground.

b) Calculate its height above the ground 1.1 seconds after it's tossed.

My solution: a) 12.0/4.3 = 2.79 m/s

b) Vx=Vox + axt = 12+4.3(1.1)=16.73 m

3) A person jogs around a circular road with a radius of 2.30 km. The jogger goes half way around in 17 minutes then stops because of a muscle cramp.

a) What is the total distance run?

b) Calculate the magnitude of the jogger's displacement.

c) Calculate the magnitude of the average velocity in m/s (C=2pir)

My solution: a) 2.30^2 + 2.30^2 = 10.58 km

b) Vx^2=Vox^2+2axX = 2.30^2+2(17)=39.29 km/min

c) C=2pir = 14.45 m/s

4) A car leaves Westown heading for Easton, 120 miles away at an average velocity of 55 mi/h east. At the same time a car pulls out of Easton at 40 mi/h heading west.

a) How long will it take for them to meet (in hrs)?

b) Find the distance in miles from Westown where the two cars cross.

My solution: a) 120/55 = 2.66

120/50 = 3

3+2.66/2=2.8 hrs

b) 55 + 40 = 95

120 - 95 = 25 miles

5) A ball is thrown straight up and reaches a maximum height of 15.3 m. At what height from the launch point is the velocity equal to one half the initial velocity?

My solution: 15.3/9.8 = 1.56

1.56/.5 = 3.12 m

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# Homework Help: Forces, magnitude, and velocity please help!

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