A rock is thrown nearly vertically upward from the edge of a tall building of he

Question

A rock is thrown nearly vertically upward from the edge of a tall building of height "H". It just misses the edge of the building on the way down and strickes the ground "T" seconds after being thrown.
Given H T
Determine initial velocity of the rock maximum height it reaches above its starting point the time to reach its maximum height



A rocket starts from rest and accelerates vertically upward at a value "a" for "T1" seconds. The engine are thrn cut off with only gravity affecting the rocket. Given a T1
Determine maximim height attained by the rocket the time from launch to just before hitting the ground the speed of the rocket just before hitting the ground

Explanation / Answer

These are the equations you would use:

v = vo + at

x = (1/2)(v + vo)t

x = vot + (1/2)at

v2 = v20 + 2ax

Now, when taking a closer look both of these are essentially the same problem - falling body. The moment the ball is thrown is equivalent to the moment the rocket reaches T1 seconds because they are both launched upward from a certain height.

To calculate the maximum vertical height, you must first find the time it takes to reach that point by using the first equation listed and setting the velocity equal to zero. Then you would plug in your initial velocity and the negative acceleration of gravity (since it is constantly being pulled downward on its way ) to solve for time:

v = vo + at turns into 0 = vo + at       which gives you       t = vo / a

You would then use this time you obtained to solve for the height by plugging it into the forth equation. Again, you would set velocity equal to zero and plug in the factors you have which are inital velocity and acceleration. You also know that the inital x value is zero since it the point from which it was thrown. Thus giving you the equation for maxium height:

x = ( v20 / 2a )

To find the time it takes to hit the ground you would use the thrid equation, setting the left side equal to the height of the building or rocket at T1. You then plug in inital velocity( which will be negative in this case due to fact that we want to know when it hits the ground) and the acceleration of gravity ( which is positive in this case since it is working towards the speed of the body before hitting the ground ). By moving the left side of the equation over to the right and setting it equal to zero, you can then use the quadratic formula to solve for the time when it hits. Your answer would be the positive outcome, not the negative.

To find the speed at which it hits the ground, simply plug the known variables into the last equation. Your intial velocity will be negative since its thrown upward, your acceleration will be positive since its gravity pulling it downward, and x will be the height of the builing:

v2 = v20 + 2ax

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