a body is thrown from the top of a tower of 180m height. At the same time another body thrown vertically upward with a velocity of 60m/s.when and where will the bodies meet?
Answers
Explanation:
Well, here is my degree on the line.
1. The masses are not important. All bodies fall at the same speed.
2. “The displacement is 0.” I am not certain, but I believe the statement means “Ignore effects of the atmosphere.”
The term “thrown” is interpreted to mean thrown sideways off the building so as not to impart any downward velocity.
3. My simple mind likes to model the motion of the two bodies rather than just solve an equation. Particularly one that invokes the solution to a quadratic equation. To simplify the math, I solved the equations graphically and found that the two curves cross at 3 seconds.
4. Body #1 starts at 180 m off the ground and 0 m/sec velocity. It accelerates for 3 seconds at 9.8 m/s every second, or 3 s x 9.8 m/s/s = 29 m/s. It has traveled downward 0.5 x 9.8 m/s x (3 s)^2 = 43 m (down) from 180m. That is 137 m in the air.
5. Body #2 is shot upward at 60m/s^2. In 3 sec it is traveling at a speed of 60m/s - (9.8m/s/s x 3s), or 18.4 m/s. It has traveled (60m/s x 3 s) - 0.5 x 9.8 m/s^2 x (3 s)^2 = 137 m in the air
6. If you plot the equations of motion you find that object #1 hits the ground in 6 seconds. Object #2 hits the peak at 6 seconds and falls to the ground in another 6 seconds. Its path is a parabola.
It is very satisfying to know that Mr. Fredericks also got 3 seconds. I have not tried to solve this problem (aka “shoot the monkey”) in over fifty years.