A cannonball is shot (from ground level) with an initial horizontal velocity of

ID: 1999489 • Letter: A

Question

A cannonball is shot (from ground level) with an initial horizontal velocity of 37 m/s and an initial vertical velocity of 24 m/s.

What is the initial speed of the cannonball?

What is the initial angle of the cannonball with respect to the ground?

What is the maximum height the cannonball goes above the ground?

How far from where it was shot will the cannonball land?

What is the speed of the cannonball 1.4 seconds after it was shot?

How high above the ground is the cannonball 1.4 seconds after it is shot?

initial horizontal speed ux = 37 m/s

initial verttical speed uy = 24 m/s

speed= sqrt(ux^2 + uy^2) = 44.10 m/s ,..............Ans

angle @ = tan^-1(uy / ux) = tan^-1(37/24) = 57.03 deg ....Ans

max. height will be when vertical velocity is zero so using

0^2 - 24^2 = 2 * -9.81*h

h = 29.36 m ........Ans

x = v^2 sin(2@) / g = 44.10^2 x sin(2*57.03) / 9.81 = 181.02 m ....Ans

only verrtical speed changes and horizontal velocity remains same.

after 1.4 sec

vy = 24 - (9.81*1.4) = 10.27 m/s

speed = sqrt(37^2 + 10.27^2) = 38.40 m/s //////........Ans

h = 24*1.4 - (9.81 x 1.4^2 / 2) = 24 m ..Ans

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