Question

A lamp is 50 ft. above the ground. If the ball is dropped from same height from a point 30 ft. away from the light pole, the ball falls at distance , s=16{t}^{2}ft. in 't' seconds, then how fast is the shadow of the ball moving along the ground ½ second later ?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A lamp is 50 ft. above the ground. If the ball is dropped from same height from a point 30 ft. away from the light pole, the ball falls at distance , s=16{t}^{2}ft. in 't' seconds.

Let assume that A be the required position of ball which is 30 ft. away from light pole and ball falls at a distance 16t^2 ft.

So,

At time t, ball drops 16t^2 ft distance, therefore

$\rm :\longmapsto\:y = 50 - {16t}^{2}$

When t = 0.5 sec,

$\rm :\longmapsto\:y = 50 - {16(0.5)}^{2} = 50 - 16 \times \dfrac{1}{4} = 50 - 4 = 46 \: ft$

➢ Let assume that light pole be represented as BC and length of shadow x be DE.

➢ Let further assume that angle BEC = k.

So,

$\red{\rm :\longmapsto\:In \: \triangle \: EAD}$

$\rm :\longmapsto\:tan \: k = \dfrac{AD}{ED}$

$\rm :\longmapsto\:tan \: k = \dfrac{50 - 16{t}^{2} }{x} - - - (1)$

Also,

$\red{\rm :\longmapsto\:In \: \triangle \: EBC}$

$\rm :\longmapsto\:tan \: k = \dfrac{BC}{EC}$

$\rm :\longmapsto\:tan \: k = \dfrac{50}{30 + x} - - - - (2)$

On equating equation (1) and (2), we get

$\rm :\longmapsto\:\dfrac{50 - 16{t}^{2} }{x} = \dfrac{50}{30 + x}$

$\rm :\longmapsto\:50 - 16{t}^{2} = \dfrac{50x}{x + 30} - - - - (1)$

When t = 0.5 sec,

$\rm :\longmapsto\:50 - 16 \times {0.5}^{2} = \dfrac{50x}{x + 30}$

$\rm :\longmapsto\:50 - 4 = \dfrac{50x}{x + 30}$

$\rm :\longmapsto\:46 = \dfrac{50x}{x + 30}$

$\rm :\longmapsto\:46x + 1380 = 50x$

$\rm :\longmapsto\:1380 = 50x - 46x$

$\rm :\longmapsto\:1380 = 4x$

$\bf\implies \:x = 345 \: ft$

Now, from equation (1), we have

$\rm :\longmapsto\:50 - 16{t}^{2} = \dfrac{50x}{x + 30}$

$\rm :\longmapsto\:50 - 16{t}^{2} = 50\bigg[\dfrac{x}{x + 30} \bigg]$

$\rm :\longmapsto\:50 - 16{t}^{2} = 50\bigg[\dfrac{x + 30 - 30}{x + 30} \bigg]$

$\rm :\longmapsto\:50 - 16{t}^{2} = 50\bigg[1 - \dfrac{30}{x + 30} \bigg]$

On differentiating both sides w. r. t. t, we get

$\rm :\longmapsto\:\dfrac{d}{dt}(50 - 16{t}^{2} ) = 50\dfrac{d}{dt}\bigg[1 - \dfrac{30}{x + 30} \bigg]$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx}k = 0}}$

$\boxed{ \bf{ \: \dfrac{d}{dx} {x}^{n} = n {x}^{n - 1} }}$

$\boxed{ \bf{ \: \dfrac{d}{dx} \frac{1}{ {x}^{n}} = - \: \frac{n}{ {x}^{n + 1} }}}$

So, using these results, we have

$\rm :\longmapsto\:0 - 32t= 50\bigg[0 + \dfrac{30}{(x + 30)^{2} } \dfrac{dx}{dt}\bigg]$

$\rm :\longmapsto\: - 32t= \dfrac{1500}{(x + 30)^{2} } \dfrac{dx}{dt}$

$\rm :\implies\:\dfrac{dx}{dt} \: = \: - \: \dfrac{32t {(x + 30)}^{2} }{1500}$

On substituting the values of x = 345 and t = 0.5, we get

$\rm :\implies\:\dfrac{dx}{dt} \: = \: - \: \dfrac{32(0.5) {(345 + 30)}^{2} }{1500}$

$\rm :\implies\:\dfrac{dx}{dt} \: = \: - \: \dfrac{16 {(375)}^{2} }{1500}$

$\bf\implies \:\dfrac{dx}{dt} \: = \: - \: 1500 \: ft \: per \: sec$