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Projectile Motion Example
 A small metal ball is thrown upwards, at 30° to the horizontal, with an initial speed of 20ms-1 Assuming that air resistance is negligible, calculate i)   the time taken by the ball to reach its highest point ii)  the maximum height reached iii) the horizontal distance moved by the ball when it hits the ground (assuming it is thrown above a horizontal surface) iv) the magnitude and direction of the velocity of the ball 1.6s after it was thrown. i) At maximum height, the vertical component of the velocity is zero; the ball is, of course, still moving but only horizontally. To find the time taken we use with v = 0 and u = 20cos60 Time to reach maximum height is t = 1.02s ii) The average value, va of the vertical component of the velocity of the ball is therefore the maximum height reached is Maximum height reached is h = 5.1m iii) Remembering that the total time in the air is 2t and that the horizontal speed is constant we have so, the range of the projectile on a horizontal plane is x = 35.3m iv) We know that the magnitude of the horizontal component does not change. So if we find the magnitude of the vertical component of the velocity at t = 1.6s, a simple vector addition will give us the answer. Considering the vertical component only: so, at t = 1.6s, the vertical component of velocity is v = 5.68ms-1 downwards. The horizontal component of the velocity is 20cos30 = 17.32ms-1 Inspection of the vector diagram here gives us and Therefore, at t = 1.6s, the ball is moving at 18.23ms-1 at 18.2° to the horizontal.
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