Question

An artillery shell is fired with an initial velocity of 300 m/s at 59.0° above the horizontal. To clear an avalanche, it explodes on a mountainside 41.0 s after firing. (a) What are the x and y components of the shell where it explodes, relative to its firing point? (b) What is the velocity of the shell (magnitude and direction) just before explosion? (c) What was the maximum height reached by the projectile? (d) How long did it take for it to reach the maximum height? What was the velocity and the acceleration of the projectile at the maximum height?

Answer 1

Horizontal component of velocity, Vx = 300 cos (59) = 154.51 ms-1

The x coordinate of the exploding shell relative to the firing point is :-
x (coord) =Vx *41
x (coord) = 6334.97m
x (coord) = 6.335 km

The y coordinate of the exploding shell relative to the firing point is :-
y (coord) =Vy*41-0.5*9.8*41^2
x (coord) = 2306.26m
x (coord) = 2.306 km

Time to reach max height:-
v = u +at
V = Vy -gt
At max height final velocity, V =0
-Vy = -gt
Time, t, to reach max height = 300 x sin(59)/9.81
t = 257.15/9.81
t = 26.213 seconds

Maximum height reached
Smax = (257.15 x 26.213)- (0.5 x 9.81 x 26.213^2)
Smax = 3370.34 m

Time to reach max height =26.213 s

Velocity at max height =Vx=154.51 i m/s

Acceleration at max height =-9.8j m/s^2