Question

A ball is thrown upward with a speed of 15 metre per second find the distance travelled by the ball in two second​

Answer 1

Answer:

Let us take the point from which the ball is thrown up as origin and upward direction as positive and downward direction as negative.

The initial velocity u with which the ball is thrown up= 15m/s

The acceleration a acting on the ball = - 10m/s² ( negative sign as it is directed downward).

Time t = 2s.

Distance travelled can be obtained using the relation; s = u t + ½ a t². Substituting various values, we get,

s = 15m/s×2s - ½ ×10 m/s² × 2² s²= 30m - 5 m/s² ×4s² = 30m - 20m= 10m.

The distance travelled by the ball in first two seconds= 10m upwards from the point of projection.

Explanation:

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Answer 2

$\green\bigstar$Answer:

• Distance = 30m

$\pink\bigstar$ Given:

• A Ball is thrown upward with a speed of 15 m/s
• Time = 2 seconds

$\red\bigstar$To find:

• The distance travelled by the ball in two seconds

$\blue\bigstar$ Solution:

We know that :

$\sf \: \boxed { \rm\: \: Distance \: = \: Speed \: \times \: Time}$

Given :

• Speed = 15 m/s
• Time = 2 seconds

$\therefore$By substituting the given values, we get :

$\sf \: Distance \: = \: (15m {s}^{ - 1} ) \: \times \: (2 \: s)$

[ By substituting the values ]

$\sf\implies \boxed{ \sf \: Distance \: = \: 30 \: m}$

$\sf \: [ \because \:( {s}^{ - 1}) \: and \:( s) \: got \: cancelled]$

$\sf \boxed{ \sf\:\therefore Distance \: = \: 30 \: m}$

$\pink\bigstar$ Concepts Used:

• Distance = Speed × Time
• Substitution of values
• Cancellation of same thing multipled and Divided
• Transposition Method

$\green\bigstar$Extra - Information:

• $\sf \: Speed\: = \: \dfrac{Distance}{Time}$

• $\sf \: Time \: = \: \dfrac{Distance}{Speed}$

$\hookrightarrow$ The three Equation of Motion are:

• $\sf \: v \: = \: u \: + \: at$
• $\sf \: s \: = \: ut \: + \: \dfrac{1}{2} a {t}^{2}$
• $\sf \: 2as \: = \: {v}^{2} \: - \: {u}^{2}$