Question

an object is thrown horizontally with an initial speed 10 m s - 1 from the top of the building of the height 100m what is the horizontal distance covered by the object

Answer 1

In vertical direction:

Height covered = 100 m

Acceleration due to gravity = 9.8 m/s^2

Initial velocity in vertical direction = 0 m/s

Let the time taken to reach the ground be t.

Using seconds equation of motion:

$h = ut + \frac{1}{2}at^2$

$100 = 0 + \frac{1}{2}\times 9.8 \times t^2$

$t = \sqrt{\frac{2\times 100}{9.8} }$

t = 4.51 seconds

Now, in the horizontal direction:

Initial velocity (v) = 10 m/s

Acceleration = 0

time taken (t) = 4.51 seconds

distance covered = s

s = velocity × times

s = 10 × 4.51

s = 45.1 m

Hence, the distance covered in the horizontal direction is 45.1 m

Answer 2

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