Question

A ball is thrown vertically upwards with an initial velocity of 30m/s. Find (i) the time taken to reach its highest point (ii) the distance then travelled (assume g=10m/s2)

Answer 1

Answer:

▪︎Question:

$\textsf{A ball is thrown vertically upwards with an initial}$$\textsf{velocity of 30m/s.Find (i) The time taken to reach}$$\textsf{its highest point . (ii) The distance then travelled}$$\textsf{(assume g=10m/s²)}$

▪︎Given:

Initial velocity of the ball that was thrown vertically:

>> 30 m/s

Gravity or acceleration affect the ball downward:

>> 10 m/s²

Velocity of the ball is:

>> 0

▪︎To find:

(i) The time taken to reach the highest point at the initial velocity of 30 m/s .

(ii) The distance travelled by the ball vertically.

▪︎Taken:

Formula to find the time taken by the ball to reach the highest point:

=> v=u+at

Where,

v=Velocity

u=Initial velocity of the ball

a=Acceleration

t=Time

( Acceleration is 9.8 m/s)

Now , formula to find the distance travelled by the ball vertically:

Let the distance travelled be x

Equation:

=> v²=u²+2ax

Where,

v=Velocity

u=Initial velocity

a=Acceleration

x=Distance travelled by the ball

▪︎Concept:

Use the formula given to find the time taken by the ball to reach the highest point and the distance travelled by the ball .

And also used cross multipllocation in the solution.

▪︎Solution:

Time taken by the ball to reach the highest point:

$\bold\purple{Taken,v=u+at}$

>> v=u+at

>> 0=30m/s+(9.8m/s)t

>> 0=30m/s+9.8t

>> -9.8t+0=30m/s

>> -9.8t=30m/s-0

>> -9.8t=30m/s

>> t=30m/s÷9.8

>> t=3.0612(approx)

____________________________________

Distance travelled by the ball vertically:

$\bold\green{Taken,v²=u²+2ax}$

>> v²=u²+2ax

>> 0=30m/s×30m/s+2(9.8)x

>> 0=900+(19.6)x

>> 0=900+19.6x

>> -19.6x+0=900

>> -19.6x=900-0

>> -19.6x=900

>> x=900/19.6

>> x=45.918 (approx)

▪︎Answer:

• So , the time taken by the ball to reach the highest point is 3.0612 second (approx) .

• And the distance travelled by the ball vertically is 45.918 meter (approx) .

▪︎Extra information:

To find acceleration:

$\bold{a=\frac{V-V1}{T}}$

Where,

a=Acceleration

V=Final velocity

V1=Initial velocity

T=Time taken

To find velocity:

$\bold{v=\frac{D}{T}}$

Where,

v=Velocity

D=Displacement

T=Time taken

HOPE IT HELPS YOU

Answer 2

Answer:

➡︎Question:

A ball is thrown vertically upwards with an velocity of 30m/s.Find (i) The timits highest point . (ii) The distance then travelled .

➡︎Given:

⭐Initial velocity of the ball that was thrown vertically:

✏ 30 m/s

⭐Gravity or acceleration affect the ball downward:

✏10 m/s²

⭐Velocity of the ball is:

✏0

➡To find:

(i) The time taken to reach the highest point at the initial velocity of 30 m/s .

(ii) The distance travelled by the ball vertically.

➡︎Taken:

Formula to find the time taken by the ball to reach the highest point:

=> v=u+at

Where,

✔v=Velocity

✔u=Initial velocity of the ball

✔a=Acceleration

✔t=Time

( Acceleration is 9.8 m/s)

Now , formula to find the distance travelled by the ball vertically:

✏Let the distance travelled be x

➡Equation:

=> v²=u²+2ax

Where,

✔v=Velocity

✔u=Initial velocity

✔a=Acceleration

✔x=Distance travelled by the ball

➡︎Concept:

Use the formula given to find the time taken by the ball to reach the highest point and the distance travelled by the ball .

➡Solution:

✏Time taken by the ball to reach the highest point:

⚫Taken,v=u+at

✔ v=u+at

✔0=30m/s+(9.8m/s)t

✔ 0=30m/s+9.8t

✔-9.8t+0=30m/s

✔ -9.8t=30m/s-0

✔ -9.8t=30m/s

✔ t=30m/s÷9.8

✔ t=3.0612(approx)

__________________________

➡Distance travelled by the ball vertically:

✔ v²=u²+2ax

✔ 0=30m/s×30m/s+2(9.8)x

✔0=900+(19.6)x

✔ 0=900+19.6x

✔ -19.6x+0=900

✔ -19.6x=900-0

✔ -19.6x=900

✔ x=900/19.6

✔ x=45.918 (approx)

➡︎Answer:

✏So , the time taken by the ball to reach the highest point is 3.0612 second (approx) .

✏ And the distance travelled by the ball vertically is 45.918 meter (approx) . ✔

Explanation:

HOPE IT HELP YOU