Problem 1: Projectile motion A cannonball is launched across a gully with an ini
ID: 2042053 • Letter: P
Question
Problem 1: Projectile motion
A cannonball is launched across a gully with an initial velocity of V(o) = 40 m/s at an angle of Theta = 60 above
the horizontal, as shown below. A survey of the gully indicates that its shape can be approximated by the
equation y(s) = ax + bx^2, with a = -0.4 and b = 0.004 m^-1, in a coordinate system whose origin is located
at the position of the cannon. Air resistance on the cannonball can be ignored.
a) What is the maximum height h of the cannonball above the cannon?
b) What are the {x,y} - coordinates of the point of impact where the cannonball hits the gully?
c) What length of time is the cannonball in the air?
d) What is the speed of the cannonball when it hits the gully?
See link above if you need the picture
Explanation / Answer
(a) ux = 40cos60 = 20m/s
uy = 40sin60 = 34.64 m/s
maximum height is reached when vy=0
v2=u2+2ah
0=34.642-2*9.81*s
h=61.162m
(b) cannonball hits the gully where ycannonball=ygully
ut - 0.5gt2 = -0.4x+0.004x2 where x=uxt = 20t
34.64t - 4.905t2 = -0.4(20t)+0.004(400t2)
42.64t-6.505t2 = 0
t = 6.555 s
x = 20(6.555) = 131.1m y=34.64(6.555)-4.905(6.5552) = 16.3m
coordinates of the point of impact = (x=131.1m, y=16.3m)
(c) t=6.555s (from part b)
(d) At the impact, vx=20m/s
vy=u-gt=34.64-9.81(6.555) = -29.66 m/s (downward)
speed = (v2x+v2y) = 35.77 m/s