Question

A boy is standing on a plank which is accelerating vertically upwards with constant acceleration. If boy throws a ball vertically up with velocity V, relative to plank and its time of flight is T, the acceleration of the plank is (2) a = Vo+g7 T T (6) 197 2V, -gT T​

Answer 1

QUESTION-:

A boy is standing on a plank which is accelerating vertically upwards with constant acceleration. If boy throws a ball vertically up with velocity V, relative to plank and its time of flight is T, the acceleration of the plank is

EXPLANATION-:

• Initial velocity(u)=v₀
• Acceleration(a)=a+g(acceleration due to gravity)

We know that-:

$\longrightarrow \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{ \rm\; \blue{ Time\;of\;flight=\frac{2u}{a} }}}}$

Putting values-:

→Time of flight(t)=2v₀/g+a

CROSS MULTIPLY-:

→a=(2v₀-g)/t

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Answer 2

Explanation:

QUESTION-:

A boy is standing on a plank which is accelerating vertically upwards with constant acceleration. If boy throws a ball vertically up with velocity V, relative to plank and its time of flight is T, the acceleration of the plank is

EXPLANATION-:

Initial velocity(u)=v₀

Acceleration(a)=a+g(acceleration due to gravity)

We know that-:

\longrightarrow \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{ \rm\; \blue{ Time\;of\;flight=\frac{2u}{a} }}}}⟶★

Timeofflight=

a

2u

Putting values-:

→Time of flight(t)=2v₀/g+a

CROSS MULTIPLY-:

→a=(2v₀-g)/t