Question

Given graph (for a projectile motion) shows variation
of vertical velocity (y-component of vector) with
time. Then the maximum height of projectile is
[g = 10 m/s21
v (m/s)
t(s)
(1) 10m
(3) 5 m
(2) 20m
(4) data is insufficient ​

Answer 1

Answer:

H = (1/2)*gt^2

Maximum height is reached when vertical component of velocity becomes zero.

That is from the graph it took 1s to reach the max height H

Using

H = (1/2)*gt^2

H = 0.5*10*1*1 = 5m

So max height reached is 5m

Answer 2

Answer:

5m

Explanation:

vertical displacement of particle is given by eq

                                      s= ut + 1/2gt²

•        and at max height :  usinθ = 0
•         at t= 1    

                             y = u sinθ t +  1/2 gt²

                              y = (0) +1/2×10×1²

                               y= 5m................simple