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
horizontal component of velocity
=75cos(55)
=43.02 m/s
vertical component of velocity
=75sin(55)
=61.44
Use the range formula to find distance in the x direction,
R=(vorigional^2 sin2(55))/g
R=(75^2sin(110))/-9.8
R=528.58 m
Therefore the total distance in the x-direction is 528.58 m
Max height will occur in the middle so
dx at Max Height = (528.58/2)
=264.28 m
time at which max height occurs =
=264.28/(75cos55)
=6.14s
dy=vosin55t+0.5gt^2
dy= 75sin(55)6.14 + 0.5(-9.8)(6.14)^2
dy=192.5m
Therefore the max height is 192.5m and it is obtained at a horizontal distance of 264.28 m
ok thnx
but this is what i did for B
i found the time using
t = 75sin(55)/ 9.81 = 6.3
(frm the formula
t= vsinØ / g
I used t (6.3) in the formula
h = ut + 1/2gt
= 75*6.3 + 1/2 (9.81) (6.3)^2
= 472.5 + 194.5
= 667 m