Question

7. A ball is dropped from the top of a building and is found to travel 50m in the last second before it reaches the ground. Find the height of the building.

Answer 1

Answer:-

151.25m

Explanation:-

Let the height of the building be h and the time be t.

Since, it is given that the ball travels 50m in the last second . Thus it is clear that in (t-1)s it will travel a distance of (h-50)m.

In this case:-

• Initial velocity of the ball will be zero

• We will take g = 10m/s² for

simplification in calculations.

By using 2nd equation of motion:-

=> h = ut + 1/2gt²

=> h = 1/2×10×t²

=> h = 5t² ----[1]

Now, when height is (h-50)m and time

is (t-1)s :-

=> h-50 = ut + 1/2×10×(t-1)²

=> 5t²-50 = 5(t-1)². [∵ h = 5t² ]

=> 5t²-50 = 5t²+5-10t

=> -55 = -10t

=> t = -55/-10

=> t = 5.5s

Now, by putting the value of t in eq.1, we get:-

=> h = 5(5.5)²

=> h = 151.25m

Thus, height of the buiding is 151.25m .