A ball is thrown out of a window 18m off the ground. It is thrown horizontally at 0.5m/s. Show that it takes about 2s to reach the ground

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A ball is thrown out of a window 18m off the ground. It is thrown horizontally at 0.5m/s. Show that it takes about 2s to reach the ground

Using the initial information we can see that the acceleration in the 'y' direction is equal to 'g', so from this we can work out the velocity in the 'y' direction 'V(y)' by simply integrating the acceleration with respect to time; so V(y) = 9.8t +c and as the ball is initially at rest c=0. By intergrating the velocity in the y direction V(y) we can then obtain the displacement in the y direction 'S(y)'. So S(y)=0.5*gt²+k. Now as we are taking the window as the origin k=0. So S(y)=0.5*gt². When the ball hits the ground the displacement it 18m or S(y)=18. Using our formula for S(y), we can solve for t. t comes out to be 1.92 to 2 d.p, which is roughly 2.

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A ball is thrown out of a window 18m off the ground. It is thrown horizontally at 0.5m/s. Show that it takes about 2s to reach the ground

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