Question

If an object is dropped from the top of building and it reaches the ground at t = 4S, then the height of the building is (ignoring air resistance; g = 9.8 ms-2). a) 77.3 m b) 78.4 m c) 80.5 m d) 79.2 m

Answer 1

Given :-

Time taken to reach the ground = t = 4 s

Acceleration due to Gravity = g = 9.8 ms-².

Concept :-

Since when a body is dropped or thrown from some particular height then it has final Velocity only, its initial velocity becomes zero or when a body is projected upward then it has initial velocity only.

But when a body is projected at some angle Ø then it has velocity in both horizontal and vertical direction i.e. x-direction and y-direction.

If the Ø is made with Horizontal then we take for horizontal CosØ and for vertical we take SinØ but If Ø is made with vertical we take CosØ and for horizontal we take SinØ.

Using Second equation of Motion,

S = 1/2 gt²

S = 1/2 × 9.8 × 4 × 4

S = 78.4 m

Hence, Option (b) is correct.

Answer 2

Answer:

The answer is 78.4

Explanation:

Given that,

Time t = 4 s

When an object is dropped from the top of a building and it reaches the ground.

Then the initial velocity u = 0

Using the equation of motion

s = ut+ 1/2 gt2.... (l)

Where, u = initial velocity

g = acceleration due to gravity

t = time

Put the value of t and g in equation (I)

s= 0+1/2 ×9.8×4×4

s =78.4m

Therefore,the height of the building is 78.4 m.

HOPE THIS ANSWER IS HELPFUL.