a shell is fired with a velocity of 75 m/s at an angle of 55 degrees to the horizontal. calculate the
* horizontal and vertical components of the velocity
* greatest height above ground attained by the shell
* horizontal distance travelled by the shell before it hits the ground
The problem is well stated, except I would add between the second and third step find the time to the max height, as that will help you find the last step.
So what is you difficulty? Use sin and cosine of the launch angle to find vertical and horizontal components. The previous link I sent you is a good sample problem on that.
I will be happy to critique your thinking or work.
It's another basic "plug 'n chug". Remember to always think of SOHCOHTOA (where in this case, 75m/s is the hypotenuse) when splitting into horizontal and vertical vectors:
The vertical component:
v/75 = sin55
so: v = 75(sin55)
The horizontal component:
v/75 = cos55
so: v = 75(cos55)
x0 = -0.5at^2 + vt + x
The greatest height above ground:
0 = -0.5(9.8m/s^2)t^2 + (0)t + h
where h = height
h = 0.5(9.8m/s^2)t^2
note that v = 0m/s because at the highest point, the shell stops.
Horizontal distance travelled:
0 = (0)t^2 + (75(cos55))t + d
where d = distance
d = (75(cos55))t
note that a = 0m/s^2 because there is no horizontal acceleration (in this case, gravity doesn't move sideways).
From the above plug and chug. When you find t from the combination of equations above, you can plug it into the second equation and get your answer. :)
On the second paragraph, the vertical height.
<<x0 = -0.5at^2 + vt + x <<--v is initial velocity here...
The greatest height above ground:
0 = -0.5(9.8m/s^2)t^2 + (0)t + h <<V=75sin55 should be inserted here, not zero.
w