Question

A ball is thrown upwards with a velocity of 10m s.Find the maximum height reached by the ball and the time taken to come back to the thrower(gravity 10m/s-2)

Answer 1

QUESTION:

A ball is thrown upwards with a velocity of 10m s.Find the maximum height reached by the ball and the time taken to come back to the thrower(gravity 10m/s-2)

Explanation:

given that,

A ball is thrown upwards with a velocity of 10m s.

so , here

initial velocity of ball = 10 m/s

since ,

it threw in the upward direction

let the maximum height at which ball will reach be h

so, let at the maximum height final velocity be v,

so,

we have ,

initial velocity (u) = 10 m/s

final velocity (v) = 0

gravitational acceleration (g) = 10 m/s²

1.( by the motion formulae of gravitation

v² = u² + 2gh

now putting the values

(0)² = (10)² + 2(10)h

20h + 10000 = 0

20h = -10000

h = -10000/20

h = -500 m

negation of height shows the height against the gravitational force .

so,

maximum height at which ball will reach = 500 m

2.)

let the time taken by ball to come back to the thrower be t

by the motion law of gravitation

v = u + gt

putting the values,

0 = 10 + 10t

10t = -10

t = -10/10

t = -1

negation of time shows time taken to go upwards

now,

total tine taken = 2(1)

= 2 seconds

the time taken by the ball to come back to the thrower = 2 seconds

Answer 2

Answer:

Explanation:

Given

Initial Velocity (u)=10 m/s

Gravity Acceleration(g)=-10 m/s^2

Final Velocity (v)= 0m/s

°.° its moving upwards

Height=?

So it is the equation to find height is

V^2 = U^2 +2gh

(0)^2 = (10)^2 +2 (-10) h

0 = 100 - 20h

-20h = 100

h = 100/-20

h = -5 m

°.° it is against earth's gravity

.°. h = 5m

To find time taken, use the 1st equation of motion

V = U + GT

0 = 10 + (-10)T

0 = 10 - 10T

-10T = 10

T = 10/-10

T = -1 s

°.° it is against earth's gravity

.°. T = 1 s

°.° Total time taken to go upwards and come back to the thrower

.°. Total Time Taken = 2 (1)

= 2 s

.°. Height = 5m

Time taken = 2s