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/s2)
V, (m/s)
(1) 10 m
(3) 5 m
(2) 20 m
(4) Data is insufficient
wit 10 mls at an​

Answer 1

Answer:

(3) 5m

Explanation:

1) According to the graph, projectile reaches the maximum height at t= 1s when thrown with initial velocity 'u' at t=0.

At max height , v= 0

Using newtons Equations,

$v=u+at\\ \\ =>0=u-g*1\\ \\ =>u=10m/s$

2)

Again,$S=ut+\frac{1}{2}at^2\\ \\=>S=10*1-\frac{1}{2}*g*1^2\\ \\=>S=10-5=5m$

Hence, Maximum height is 5m.

Answer 2

Answer:

5m

Explanation:

take the answer in the pic ......

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