Question

An object projected vertically up from the top of
a tower took 5 s to reach the ground. If the aver-
age velocity of the object is 5 m s-, find its average
speed. (Given, g = 10 m s-).​

Answer 1

Answer:

13m/s

Explanation:

Height of tower= displacement of an object= verage velocity × time=5×5=25m,

Let initial velocity of the object is u.

Using s=ut+ 1/2 at² ,⇒−25=u×5− 1/2g x 5²

⇒u=20m/s ,

Now from the tower,Let maximum height reached by object is h

At maximum height velocity v = 0

using v²= u²+2as===> 0 =20²- 2gh ==>h= 20m

therefore, total distance= 2h + 25 = (2 x 20) + 25= 65m

Average speed= Total distance / Total Time

==>Avg.speed= 65/5= 13m/s

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