Question

1) An object is dropped from a height H. During the final second of its fall, traverses a distance of 46.5 m. What was H? m Submit

Answer 1

The object is dropped from a height so it is a free fall. So initial velocity is u0 = 0 m/s and acceleration of the body is given by acceleration due to gravity.

So a = g

At any instant of time velocity of the body is given by the equation V(t) = u0+ at. where t is the total time taken to reach from the velocity u0 to V(t).

Now in the question given that the last second of travel the object covers 46.5 m.

Assume tn is the time taken to reach the bottom. we can take tn- t(n-1) =1sec

Now lets assume V(tn) is the velocity at tn timeof travel and V(tn-1) is the veloctiy of t(n-1)

using the formula V^2 - U^2 = 2 *a*S

V - Final velocity U - starting velocity S - displacement occured during the velocity change

a - acceleration

we have V(tn)^2- V(tn-1)^2 = 2* g*S

Now we also know the V = U+ at where t -time taken to reach from initial velocity U final velocity V

V(tn) = V(tn-1)+ a*t here t=1Second and a = g putting this equation in the one equation above

(V(tn-1)+g)^2 - V(tn-1)^2 = 2* g*S

Simplifying this we will get V(tn-1) = S - 1/2*g

Where S is 46.5 m and g= 9.8. and putting this values you wil get V(tn-1)=41.6 m/s

Now timw taken the object to reach this velocity from zero initial velocity is given by

V(tn-1) - u0 = a*t

t = V(tn-1)/g = 4.245 S

total time of travel of the object to reach the bottom = 4.245 + 1 = 5.245 s

total distance travelled by the object under free fall during this time period is given by

S = uo*t + 1/2 * a *t^2

u0 = 0 m/s

a = g= 9.8

t= 5.245

S = 134.7 m

Height of the building is 134.7m