Question

OR
A particle is thrown vertically upward with velocity 10 m/s and it moves
freely under gravity find
[1+1]
(a) Time to reach at maximum height.
(b) Maximum height
10 m/s
g = 10 m/s​

Answer 1

A) v = u + at.
0 = 10 - 10*t.
-10 = -10*t.
t = 1 second
Acceleration is used as negative because the object is moving up and the gravity is pulling it downward. So negative sign is used.

B) S = ut + 1/2at*t
H = 10*1 - 1*10/2 (1*1)
H = 5 m.
In this (B) part the acceleration is taken negative for the same reason as of the (A) part.

Answer 2

Answer:

Explanation:

Given

Initial velocity of throw $u=10\ m/s$

At maximum height velocity of particle is zero

using

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$0-(10)^2=2\times (-9.8)\times (s)$

$s=\dfrac{100}{2\times 10}$

$s=5\ m$

time taken to reach maximum height

$v=u+at$

$0=10-10\cdot t$

$t=1\ s$