i just need some corrections with my solution

Gosh, Gwiyomi, you sure must like doing arithmetic.

 

I make t = 2.153 seconds and x = 20.76 metres

 

So pretty close to your results.

 

For your information about projectile problems,

 

The path of the projectile is fixed by the launch speed and angle alone.

 

If you launch up and to the right as positive (usual x and y axes) then the position at any time t is given by a pair of equations

 

x = Vcos(a)t

 

y = -1/2gt² + Vsin(a)t

 

where a is the launch angle, and V is the launch speed.

 

That is the x and y coordinates at any time in the flight are given by those two equations.

 

It is only a couple of simple lines to derive these so you should not try to remember them, just how to get them.

 

If you enter your value of y = 2 (and V and a) into the second equation it yields a quadratic in t.

The two solutions correspond to going up and coming down.

 

Knowing t you can substitute into the first equation to find the x distance at this time.

Edited February 11, 201411 yr by studiot

oh i solved according to what i understand. i usually solve the simple ones first, so that i could determine which equations should i use

 

how about letter c?

part c?

 

Well if this was an exam I wouldn't get may marks since I didn't read the question properly.

 

Red face.

 

You have correctly identified that velocity is a vector and may be resolved into horizontal and vertical components.

Further this is also true of acceleration and distance.

There is only one acceleration, that due to gravity which is vertical and downwards.

There is no horizontal acceleration (ie it is zero horizontally)

 

 

So we used the formula s = ut + 1/2 f t²

 

to solve parts a and b.

 

Note I have used s for general distance, and f for acceleration. This allow the use of a, b, c for angles instead of Greek letters and also allows us to put in the distance in the appropriate direction.

 

To solve part c we introduce a second formula

 

v = u + ft where v is the velocity at time t and u is the initial velocity.

 

Horizontally V_x = 15cos(50) + 0*t = 9.6418 m/s

 

Vertically V_y = 15sin(50) + (-9.81)*(2.153) = -9.630 m/s ie downwards.

 

Total velocity V = sqr(V_x² + V_y²) = 13.627 m/s

 

You got all this but I see you were unsure about the angle denoting the direction of travel.

 

This angle, call it b, is in the fourth quadrant since the travel is down and to the right.

 

We obtain a fourth quadrant angle by calculating a first quadrant angle © from arctan(V_y/V_x), ignoring any signs

and subtracting from 360.

 

b = 360 - c

 

This is consistent with the velocity diagram I have shown superimposed on top of the trajectory diagram below.

 

Have I completed the job this time?

 

aah thanks!!!!!

your very helpful