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question #1. A rocket is fired at a speed of 75.0m/s from ground level, at an angle of 60 degrees above the horizontal. The rocket is fired toward an 11.0m high wall, which is located in 27.0 m away. the rocket attains its launch speed ina negligibly short period of time, after which its engines shut down the rocket coasts. By how much does the rocket clear the top of the wall?

Question # 2. A marble is thrown horizontally with a speed of 15m/s the tope of a building. when it strikes the ground, the marble has a velocity that makes an angle of 65 degrees with the horizontal. From what height above the ground was the marble thrown?


can you please help me... detailed? please.......

Answer 1

First some advice then a solution. What makes a problem like this easier is if you think of the velocities in the question as the sum of velocities in the horizontal direction and velocities in the vertical direction added together. If you take them apart you can deal with each one individualy.

Assuming no wind resistance (typical with this kind of problem) horizontal velocities don't change. The rocket will just keep moving at the same velocity until acted on by another force.

To answer the problem we need to know the height (vertical) of the rocket the moment it passes over the wall. To find that we need to know how long it will take before the rocket is directly over the wall.

Draw a triangle with the rocket's path as the hypotenuse at a 60 degree angle.

The vertical velocity is 75 sin 60 m/s
The horizontal velocity is 75 cos 60 m/s

Using one of your formulas you can calculate the time it will take to get to the wall:

v = d / t
v * t = d
t = d / v
=27 m / 75 cos 60 m/s
=0.72 sec

Now the question is how high is the rocket after 0.72 sec?

We know that the rocket is travelling up at speed of 75 sin 60 m/s and it is accelerating down at 9.8 m/s^2. We'll choose to make that negative because it has to be the opposite sign as the velocity (gravity makes things slow down when travelling up). So we need another one of our equations:

d = vi t + 1/2 a t^2

where vi (initial velocity) is 75 sin 60 m/s
t is 0.72 sec
a is -9.8 m/s^2

d = (75 sin 60 m/s) (0.72 sec) + 1/2 (-9.8 m/s^2) (0.72 sec)^2
=44.2 m

So if the wall is 27 m tall then the rocket clears the wall by (44.2m - 27m= 17.2m). You should write your answer as 17 m because of the accuracy level of the information you are given (height of the wall).

A good place for formulas is: http://nas.cl.uh.edu/blanford/FormulasMechanics.htm

A solution to a question like #2 can be found at:
http://tutor4physics.com/example.htm

Remember, physics is like learning a new language. When you're getting started you need practice, practice, practice. Good luck!

Answer 2

First find the vertical (y) and horizontal (x) components of the rocket's initial velocity; let this velocity be v, and the angle ø. Then

vy = v*sinø and vx = v*cosø

The horizontal velocity will remain constant throughout the flight, and the rocket will travel horizontally a distance sx(=27m) to the wall. The time it takes for the rocket to reach the wall location is then t=sx/vx.

The vertical position above ground of the rocket is given by y=vy*t-.5*g*t^2. Plug in t from the horizontal calculation into this to get y at the wall location. The clearance is then y - wall height.


The angle of arrival of the marble is the arctan(vy/vx), where vy and vx are the vertical and horizontal velocity components at the time the marble hits the ground. As above, the horizontal velocity does not change, it will stay at it's initial value vx. The vertical velocity is that of a falling object, or vy=g*t. From the formula s=.5*a*t^2, the distance a falling object falls is h=.5*g*t^2. The time to fall a distance h is then t=√[2h/g]. Plug this t into the formula vy=g*t to get vy on impact, vy=g*√[2*h/g]=√[2*g*h]. vy/vx is then √[2*g*h] / vx = tan(65º) 2*g*h = vx^2*tan^2(65º), and h = vx^2*tan^2(65º) / (2*g).