Question

xample 10.3 An object is throwi
vertically upwards and rises to a height
of 10 m. Calculate (i) the velocity with
which the object was thrown upwards
and (ii) the time taken by the object to
reach the highest point.
olution: using 2 Nd equation​

Answer 1

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Answer 2

$\mathcal{\green{\underline{\blue{GIVEN:}}}}$

• Height/distance (s) = 10m.
• Initial velocity (v) = 0m/s⁻¹. [The object is thrown and then it stops momentarily]
• Acceleration due to gravity (g) = 9.8ms⁻². In other words, we are taking a= - 9.8 ms⁻² [The acceleration is against gravity, so it is negative.]

$\mathcal{\green{\underline{\blue{TO \:\: FIND:}}}}$

• Final velocity (v) = ?
• Time taken to reach the final velocity (t) = ?

$\mathcal{\green{\underline{\blue{SOLUTION:}}}}$

Recall the three equations of motion:

$\to \sf{\red{v=u+at}}\\\\\to \sf{\red{s=ut+\frac{1}{2}at^2 }}\\\\\to \sf{\red{2as=v^2-u^2}}$

It is apt to use the third formula for the final velocity and first formula for the time.

$\to \sf{\red{2as=v^2-u^2}}$

Substitute the values.

$\to \sf 2 \times (-9.8) \times 10 = 0^2-u^2$

$\sf \to -196=-u^2$

$\to \sf u=\sqrt{196}$

$\implies\boxed{\sf{\blue{u=14m}}}$

Now use the first equation.

$\to \sf{\red{v=u+at}}$

Substitute the values.

$\to \sf 0=14+(-9.8) \times t$

$\to \sf t=\frac{-14}{-9.8}$

$\implies\boxed{\sf{\blue{t \approx 1.43 \:\: seconds.}}}$

$\mathcal{\green{\underline{\blue{ANSWER:}}}}$

• Distance = 14m.
• Time ≈ 1.43 seconds. [≈ means approximately. In other words, time is approximately 1.43 seconds]

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