A large cannon is fired from ground level over level ground at an angle of 19 degrees above the horizontal. This initial speed is 202 m/s. Neglecting air resistance

Question

how far will the projectile travel in the horizontal direction before striking the ground?

Answer 1

angle (θ) = 19º

speed (v) = 202 m/s

horizontal component = vcosθ

vertical component = vsinθ

a = -g [negative because it's travelling downwards, so down is negative]

kinematic equation used to find time [time in horizontal direction is equal to time in vertical direction:

x = vt + 1/2at²

0 = vsinθt - 1/2gt²

time of flight = 2vsinθ

range of flight = [(vcosθ)(2vsinθ)/g] = [(202cos(19º)(2*202sin(19º))/(9.8)] = 2563.4 m = 2.56 km

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