A cannon in a fort is 18.4 m above the ground, and it fires a cannonball with an

Question

A cannon in a fort is 18.4 m above the ground, and it fires a cannonball with an initial velocity of 72.4 m/s at a 34 degree angle above the horizontal. How far from the base of the fort wall does the cannonball strike the ground? The method solve this problem is to calculate the time of flight using d = 1/2 at^2 + v_i t, using the vertical velocity for v_i, and solve for t with the quadratic equation (or by completing the square). Then solve for distance in the usual manner using the horizontal velocity.

Explanation / Answer

Given,

h = 18.4 m ; v0 = 72.4 m/s ; theta = 34 deg

v0x = v0 cos(theta) = 72.4 x cos34 = 60.02 m/s

v0y = v0 sin(theta) = 72.4 x sin34 = 40.49 m/s

The max height reached by cannon will be:

h = v0y^2/2g = 40.49^2/2 x 9.8 = 83.64 m

So the total height above the ground will be:

H = 18.4 + 83.6 = 102 m

the time taken to fall freely will be:

t = sqrt (2h/g) = sqrt(2 x 102/9.8) = 4.56 s

time taken to cover h,

t = v0y/g = 40.49/9.8 = 4.13 s

T = t + t' = 4.56 + 4.13 = 8.69 s

D = v0x t = 60.02 x 8.69 = 521.57 m

Hence, D = 521.57 m

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