Question

A ball falls of a table and reaches the ground in 1 second. Assuming g=10m/s2, Calculate it's speed on reaching the ground and the height of the table.​

Answer 1

Initial velocity u=0 Let final velocity is v and height is h.

Using v=u+at

⇒v=0−10×1=−10m/s

Using s=ut+21at2

⇒−h=0×1−21×10×12  ⇒h=5m

Answer 2

$\\ \frak{Given} \begin{cases} \bullet\: \sf initial \: velocity(u) = 0 \: m {s}^{ - 1} \\ \bullet \: \sf accelaration(a) = 10 \: m {s}^{ - 2}\\ \bullet \: \sf time \: taken(t) = 1 \: sec \end{cases}$

initial velocity was supposed to be (0) because the ball is falling from the table which means it was initially at rest.

We are asked to calculate :-

• Final Velocity
• Height of the table i.e distance traveled by the ball

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By Using 1st kinematical equation:-

$\\ \leadsto\large \underline{ \boxed{ \sf v = u + at}} \bigstar$

$\\ : \implies \sf v = 0 + 10 \times 1$

$\\ : \implies \boxed{ \boxed{\frak{ v = 10 \: m {s}^{ - 1}} }} \bigstar$

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Now, by using 2nd kinematical equation:-

$\\ \leadsto \large\underline{ \boxed{ \sf s = ut + \frac{1}{2}a {t}^{2} }} \bigstar$

$\\ : \implies \sf s = 0 \times 1 + \frac{1}{2} \times 10 \times {1}^{2}$

$\\ : \implies \boxed{ \boxed{ \frak{ s = 5 \: m}}} \bigstar$

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Note :- You can use 3rd kinematical equation as well to find the height of the table. This Can be explained as under :-

$\\ \leadsto \large\underline{ \boxed{ \sf {v}^{2} - {u}^{2} = 2as }} \bigstar$

$\\ : \implies \sf {(10)}^{2} - {(0)}^{2} = 2 \times 10 \times s$

$\\ : \implies \sf 100 = 20s$

$\\ : \implies \sf s = \cancel\dfrac{100}{20}$

$\\ : \implies \boxed{\boxed{ \frak{ s = 5 \: m}}} \bigstar$

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