a) What is the cannonball\'s velocity at the highest point in the trajectory? b)

Question

a) What is the cannonball's velocity at the highest point in the trajectory?

b) How long does it take to get to this point?

c) How far does it travel horizontally right before it hits the ground?

d) What is the magnitue and direction of the cannonball's velocity right before it hits the ground?

must use the following equations to derive answers

What is the cannonball's velocity at the highest point in the trajectory? How long does it take to get to this point? How far does it travel horizontally right before it hits the ground? What is the magnitude and direction of the cannonball's velocity right before it hits the ground? x = x0 + v0t + 1111/2at2 v = v0 + at v2 = v02 + 2a deltax

Explanation / Answer

a) Horizontal velocity component = (cos 20 x 90) = 84.57m/sec. horizontally, but 0m/sec. vertically.
Initial vertical velocity component = (sin 20 x 90) = 30.78m/sec.
b) Time to max. height = (v/g) = 30.78/9.8, =3.141 secs.
c) (3.141 x 2) x 84.57 = 531.27metres.
d) 90m/sec., and direction = 20 degrees below horizontal.

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