Question

A body is projected vertically upwards with a speed 100 m/s from the ground then distance it covers in its last second of its upward journey in meters is

Answer 1

The distance travelled by the body on last second of its upwards journey is 4.935m

What is projectile motion?

A projectile is an object or particle that is launched toward the surface of the Earth and moves along a curved path only under the influence of gravity.

To solve this question, consider the horizontal distance to be H and

distance traveled by the body in the last second be d

$H=U^{2} /2g$

$H=100^{2} /2*9.8$

H=510.2m

Now time taken to reach a height of

510.2 m= $\sqrt[n]{2H/g}$

=10.2 seconds

Now,

$s=ut+1/2gt^{2}$

$H-d= (100)(t-1)-1/2g(t-1)^{2}$

$510.2-d=100*9.2-1/2*9.8*9.2^{2}$

d=4936m

So, the correct answer is '4.936m'.

To know more about projectile motion visit:

brainly.com/question/11049671

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