Math, asked by Anonymous, 1 year ago

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 ?


✔️✔️ quality answer needed✔️✔️

❌❌NO SPAMMING❌❌​

Answers

Answered by swethatheperumal
5

ANSWER:

Draw a figure to represent the situation when the ball has fallen a distance s feet from its initial position. Let x be the 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.

Answered by Anonymous
4

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 s

We also have ds/dt = 32t and we are required to find dx/dt when

t = 1/2

Multiplying 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/sec

So 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.

Attachments:

ayusi74: hi
Anonymous: hi
ayusi74: hi
ayusi74: how are you
ayusi74: hlo
Anonymous: fine
ayusi74: yr
ayusi74: bolo
Anonymous: nothing
ayusi74: by
Similar questions