Question

Example 10.3 An object is thrown
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.
​

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Initial\:velocity=10\sqrt{2}\:m/s}}$

${\bold{\therefore Time\:taken=\sqrt{2}\:sec}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about an object is thrown vertically upwards and rises to a height

an object is thrown vertically upwards and rises to a heightof 10 m.

• We have to Calculate (i) the velocity with which the object was thrown upwards and (ii) the time taken by the object to reach the highest point.

$\underline \bold{Given : } \\ \implies Distance = 10 \: m \\ \\ \implies Acceleration(a) = - 10 \: m /{s}^{2} \\ \\ \implies Final \: velocity(v) = 0 \: m/s \\ \\ \underline \bold{To \: Find : } \\ \implies Initial \: velocity(u) = ? \\ \\ \implies Time \: taken = ?$

• According to given question :

$\bold{Using \: second \: equation \: of \: motion : } \\ \implies {v}^{2} = {u}^{2} + 2as \\ \\ \implies {0}^{2} = {u}^{2} + 2 \times ( - 10) \times 10 \\ \\ \implies \cancel - {u}^{2} = \cancel - 200 \\ \\ \implies u = \sqrt{200} \\ \\ \bold{\implies u = 10 \sqrt{2} \: m/s} \\ \\ \bold{Using \: first \: equation \: of \: motion : } \\ \implies v = u + at \\ \\ \implies 0 = 10 \sqrt{2} - 10 \times t \\ \\ \implies t = \frac{\cancel{10} \sqrt{2} }{\cancel{10}} \\ \\ \bold{\implies t = \sqrt{2} \: sec}$

Answer 2

Answer:

$\bold{\to u = 10 \sqrt{2} \: m/s}$

$\bold{\to t = \sqrt{2} \: sec=1.41\:sec}$

Step-by-step explanation:

$\to {v}^{2} = {u}^{2} + 2as \\ \\ \to - {u}^{2} = 2 \times - 10 \times 10 \\ \\ \to u = \sqrt{200} \\ \\ \bold{\to u = 10 \sqrt{2} \:m /s} \\ \\ \bold{for \: time \to} \\ \to v = u + at \\ \\ \to 0 = 10 \sqrt{2} - 10 \times t \\ \\ \to t = \frac{10 \sqrt{2} }{10} \\ \\ \bold{\to t = \sqrt{2} \: sec}$