Question

A ball is thrown vertically upwards and rises to a height of 126.5 m. Calculate the


(i) velocity by which the ball was thrown upwards and


(ii) the time taken by the ball to reach the highest point. ​

Answer 1

Given

S=126.5m

a=-9.8m/s²

v=0

v²-u²=2as

v²-0=2(-9.8)(126.5)

v²=-2479.4

a) v=49.7m/s{approx}

v=u+at

v-u=at

t=v-u/a

t=49.7-0/-9.8

Approx, 5.07sec

Answer 2

Given :-

A ball is thrown vertically upwards and rises to a height of 126.5 meters .

Required to find :-

• Velocity by which the ball was thrown upwards

• The time taken by the ball to reach the highest point

Equations used :-

v² - u² = 2as

v = u + at

Solution :-

Given information :-

A ball is thrown vertically upwards and rises to a height of 126.5 meters .

we need to find ;

The velocity by which the ball was thrown upwards ?

The time taken by the ball to reach the highest point ?

So,

From the given data we can conclude that ;

Final velocity of the ball ( v ) = 0 m/s

Displacement ( s ) = 126.5 meters

Since,

The body the is moving upwards we need to taken the acceleration due to gravity in negative .

Acceleration ( a ) = - 9.8 m/s²

Using the equation of motion ;

☞ v² - u² = 2as

☞ ( 0 )² - u² = 2 x - 9.8 x 126.5

☞ 0 - u² = 2 x - 9.8 x 126.5

☞ - u² = - - 9.8 x 253

☞ - u² = - 2,479.4

☞ Multiply with ( - ) minus on both sides

☞ - ( - u² ) = - ( - 2, 479.4 )

☞ u² = 2, 479.4

☞u = √2, 479.4

☞u = 49.79 m/s ( = 50 m/s approximately )

Hence,

Initial velocity of the ball ( u ) = 49.79 m/s

Using the next equation of motion ;

☞ v = u + at

☞ 0 = 49.79 + ( - 9.8 ) x t

☞ 0 = 49.79 + ( - 9.8t )

☞ - 49.79 = - 9.8t

☞ - ( minus ) get cancelled on both sides

☞ 49.79 = 9.8t

☞ 9.8t = 49.79

☞ t = 49.79/9.8

☞ t = 4979/980

☞ t = 5.08 seconds ( approximately )

Hence,

Time taken by the ball ( t ) = 5.08 seconds

Therefore ,

⭐ The velocity by which the ball was thrown upwards is 49.79 m/s

⭐ The time taken by the ball to reach the highest point is 5.08 seconds