A lamp is 50 feet above the ground . A ball is dropped from the same height at a point which is at a distance of 30 feet away from the light pole. If ball falls a distance s=16t² feet in t seconds . Then how fast is the shadow of the ball moving along the ground ½ s later?

 

Draw a figure to represent the situation when the ball has fallen a distance s feet from its initial position. Let x = distance along the ground from the point immediately below the ball to the point where the shadow of the ball meets the ground. We then have two similar triangles and can write down the ratio of corresponding sides in the form: x 30 ------- = ------- 50-s sWe also have ds/dt = 32t and we are required to find dx/dt when t = 1/2Multiplying out we get xs = 1500 - 30s and differentiating x(ds/dt) + s(dx/dt) = -30(ds/dt)at t=1/2 ds/dt = 16 s = 16(1/4) = 4 and 4x = 1500 - 120 x = 345 ft So 345(16) + 4(dx/dt) = -30(16) 4(dx/dt) = -6000 dx/dt = -1500 ft/secSo at this moment the shadow is moving at a speed 1500 ft/sec towards the point immediately below the ball. The high speed is to be expected since initially the shadow is at infinity and has to move to the end point in 1.76 seconds.

 

Regards

Arun (askIITians forum expert)