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=16t 2 feet in t seconds . Then how fast is the shadow of the ball moving along the ground ½ s later?

							
Dear Raghav
 
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)