Question

a boy throws a ball upwards with vel 9.8 how high does it go and how long it was in air

Answer 1

initial velocity of bol = 9.8 m/s
final velocity will be 0 after going to the highest point.

acceleration due to gravity = -9.8 m/s². (-9.8 will be because the velocity of a body is decreasing as it is going against the gravitation)

time taken by the ball to go to the highest point= final velocity - initial velocity / acceleration due to gravity

time taken = 0 - 9.8 / - 9.8
time taken = 1 seconds.

using second equation of uniformly accelerated motion,
we have,

h = ut + 1/2gt².
h = 9.8 - 4.9×1²
h = 9.8 - 4.9
h = 4.9 metres.

we can also use third equation of uniformly accelerated motion.

that is,
v² = u² + 2gh
0 = 9.8² + 2×-9.8×h
0 = 96.04 - 19.6h
19.6h = 96.04
h = 96.04/19.6
h = 4.9 metres.

the ball thrown by that boy will go 4.9 metres against the gravity.

time taken by the ball to go at the highest point = 1 second
time taken by the ball to come to ground = 1 second

time taken by ball to go at the highest point and come to the initial position = 2 seconds.

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