A cannonball a launched across a gully with an initial velocity of v0 = 40 m/s a
ID: 2042263 • Letter: A
Question
A cannonball a launched across a gully with an initial velocity of v0 = 40 m/s at an angle of theta0 = 60degree above the horizontal, as shown bellow. A surrey of the gully indicates that its shape can be approximated by the equation ys = ax + bx2, with a = - 0.4 and b = 0.001 m-1 , in a coordinate system whose origin is located at the position of the cannon. Air resistance on the cannonball can be ignored. What is the maximum height h of the cannonball above the cannon? What are the {x, y}-coordinates of the point of impact where the cannonball hits the gully? What length of time is the cannonball in the air? What is the speed of the cannonball when it hits the gully?Explanation / Answer
(a) vx = 40cos60 = 20m/s
vy = 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
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.