A ball is fired from a cannon at 20m/s at an angle of 56degrees to the horizontal. Calculate the horizontal distance the ball travels as well as its maximum height reached.
This is a typical mechanics question in an AS/A-level exam and uses the principles of SUVAT and projection.
- We can try to calculate the maximum height reached first using the veritical components.
SUVAT: s=? u=(20xsin56) v=0 a=-9.81 t=?
(v^2 = u^2 + 2as) -- 0 = (20xsin56)^2 + (2x-9.81)(s) rearange --> 274.921 = 19.62(s) s = 14.01m
- We now need to calculate the distance that the ball travels and so we must calculate the time that the ball is in the air and we can do this by using the vertical component of the ball to travel to its heighest point and then back down to the ground again. (At the top of the projection, the ball will have a vertical velocity of 0m/s)
SUVAT: s=14.01 u=20xsin56 v=0 a=-9.81 t=?
s = 0.5(u+v)xt --> 14.01 = 0.5x(20xsin56)xt t = 1.6899 (however this is the time take to reach the peak of the projection. The total time of travel will be double this as it has to come down too.) t = 3.3798
- Now that we have the total time we can use: s = ut + 0.5at^2 to find the total horizontal distance travelled.
s=(20xcos56)x(3.3798) + (0.5x0x(3.3798)^2) s = 37.799
