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

Question

A cannonball is shot (from ground level) with an initial horizontal velocity of 36.0 m/s and an initial vertical velocity of 21.0 m/s. /// 1) What is the initial speed of the cannonball? //// 2) What is the initial angle ? of the cannonball with respect to the ground? //// 3) What is the maximum height the cannonball goes above the ground?//// 4) How far from where it was shot will the cannonball land?//// 5) What is the speed of the cannonball 2.7 seconds after it was shot? //// 6) How high above the ground is the cannonball 2.7 seconds after it is shot?////

Explanation / Answer

1) Use the Pythagorean theorem. Initial Speed = Vo = sqrt [Vx^2 + Vy^2] 2) Launch Angle = arctan(Vyo/Vx) 3) g*Hmax = Vyo^2/2 where Vyo is the initial vertical velocity component. 4) Range = 2*(Vo^2/g)*sinAcosA 5) Use the Pythagorean theorem again, but with the changed value of Vy. Vy = Vyo - g*t 6) Height = Vyo*t - (g/2)*t^2

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