Question

a ball is thrown upward from the ground at initial velocity u .The ball is at the height of 80 m at two times .The time interval being 6 sec .find u​

Answer 1

Answer:

Velocity at which ball was thrown

= 50 m/s

Step by step explanations :

given that

a ball is thrown upward from the ground at initial velocity u,

here,

initial velocity of the ball = u

also given,

.The ball is at the height of 80 m at two times

and,

The time interval being 6 sec

now,

let consider that the ball is at height of

80 meter

let the velocity of the ball at the height of 80 m be ū

here,

given time interval for coming to the same height = 6 seconds,

and,

we know that,

the time taken by an object to reach the maximum height is equal to the time taken by it to reach at its initial position.

so,

here,

time taken by the ball to reach maximum height = 3 seconds,

here,

final velocity of the ball = 0 m/s

now,

we have,

initial velocity(u) = ū

final velocity(v) = 0 m/s

time taken(t) = 3 seconds

gravitational acceleration(g) = -10 m/s²

by the gravitational equation of motion,

v = u + gt

putting the values,

0 = ū + (-10)(3)

ū = 30 m/s

also,

v² = u² + 2gh

here,

h is the height attained by the ball after the 80 m

again putting the values,

(0)² = 30² + 2(-10)h

-20h = -900

h = -900/-20

h = 45 m

so,

Height attained by the ball after 80 m

= 45m

so,

Total maximum height reached by the ball = 80 + 45

= 125 m

let the velocity by which ball was thrown be u

now,

we have,

initial velocity = u

now,

we have,

initial velocity(u) = u

final velocity(v) = 0 m/s

height(h) = 125 m

gravitational acceleration(g) = -10 m/s²

by the gravitational equation of motion,

v² = u² + 2gh

putting the values,

0² = u² + 2(-10)(125)

u² - 2500 = 0

u² = 2500

u = 50 m/s

so,

Velocity at which ball was thrown

= 50 m/s

Answer 2

Answer:

initial velocity of ball = 50 m/s

Explanation:

given

at height 80 m time taken by ball to reach twice = 6 s

so

time taken to reach maximum height from 80 m = 3s

final velocity = 0 m/s

v = u + gt

0 = u - 10(3)

u = 30 m/s

h = ut + ½ gt²

h = 30(3) + ½ × (-10)(3)²

h = 90 - 45

h = 45 m

maximum height for ball = 80 + 45

125 m

v² = u² + 2gh

0 = u² + 2(-10)(125)

u² = 2500

u = 50

so,

initial velocity of ball = 50 m/s