On level ground a shell is fired with an initial velocity of 39.0 m/s at 62.0 ab

Question

On level ground a shell is fired with an initial velocity of 39.0 m/s at 62.0 above the horizontal and feels no appreciable air resistance.

A. Find the horizontal and vertical components of the shell's initial velocity. v0h, v0v =

B. How long does it take the shell to reach its highest point?

C. Find its maximum height above the ground.

D. How far from its firing point does the shell land?

E. At its highest point, find the horizontal and vertical components of its acceleration.

F. At its highest point, find the horizontal and vertical components of its velocity.

Explanation / Answer

(A)
Vh = 39.0 * cos(62) m/s
Vh = 18.31 m/s

Vy = 39.0 * sin(62)
Vy = 34.4 m/s

(B)
At highest point, Vertical velocity = 0
v = Vy + a*t
0 = 34.4 - 9.8 * t
t =  3.51 s

(C)
Vf^2 = Vy^2 + 2*a*s
0 = 34.4^2 - 2*9.8*s
s = 60.4 m

(D)
Total time of flight, = 2*3.51 = 7.02 s
Horizontal distance covered = 18.31 * 7.02 = 128.5 m

(E)
At its highest point,
the horizontal component of its acceleration, = 0
the vertical component of its acceleration, = 9.8 m/s^2

(F)
At its highest point,
the horizontal component of its Velocity, = 18.31 m/s
the vertical component of its Velocity, = 0 m/s

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