1.

A stone is thrown vertically upwards with an initial velocity of 30 m/sec. Find the maximum height it reaches and the time taken by it to reach the height.(g = 10 m/s2).

Answer»

Figure regards question:

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Understanding the question: This question SAYS that there is a stone that is thrown vertically UPWARDS with an initial velocity of 30 m/sec. We are asked to take acceleration DUE to gravity (g) as 10 m/sec sq. And we are asked to find out the maximum height it reaches and the time taken by it to reach the height.

Provided that:

  • Initial velocity = 30 mps
  • g = -10 mps sq.
  • Final velocity = 0 mps

Don't be confused!

✴️ Final velocity cames as zero because the stone is thrown upwards and at the highest POINT the final velocity will be zero.

✴️ We are taking g as negative because the stone is thrown vertically upward. It thrown upwards that is against gravity.

To calculate:

  • Maximum height
  • Time taken

Solution:

  • Maximum height = 45 m
  • Time taken = 3 sec

Using concepts:

✴️ We can use either first equation of motion or acceleration formula to calculate the time taken.

  • Choice may vary, yours!

✴️ To calculate maximum height we can use either SECOND equation of motion or third equation of motion.

  • Choice may vary, yours!

Using formulas:

✴️ The three equations of motion are mentioned below respectively:

\begin{gathered}\boxed{\begin{array}{c}\\ {\pmb{\sf{Three \: equations \: of \: motion}}} \\ \\ \sf \star \: v \: = u \: + at \\ \\ \sf \star \: s \: = ut + \: \dfrac{1}{2} \: at^2 \\ \\ \sf \star \: v^2 - u^2 \: = 2as\end{array}}\end{gathered}

✴️ The acceleration formula is mentioned below:

  • {\small{\underline{\boxed{\pmb{\sf{a \: = \dfrac{v-u}{t}}}}}}}

Where, a denotes acceleration, v denotes final velocity, u denotes initial velocity and t denotes time taken.

Required solution:

✡️ Firstly by using first equation of motion let us find out the time taken by the stone to reaches the maximum height!

{\sf{:\implies \vec{v} \: = \vec{u} + \vec{a}t}}

{\sf{:\implies 0 = 30 + (-10)t}}

{\sf{:\implies 0 = 30 - 10t}}

{\sf{:\implies 0 - 30 = -10t}}

{\sf{:\implies -30 = -10t}}

{\sf{:\implies 30 = 10t}}

{\sf{:\implies t \: = 3 \: sec}}

{\sf{:\implies Time \: taken \: = 3 \: sec}}

  • Therefore, time taken to reaches the maximum height = 3 sec!

✡️ By using acceleration formula let us find out the time taken now!

:\implies \sf Acceleration \: = \dfrac{dv}{dt} \\ \\ :\implies \sf Acceleration \: = \dfrac{\Delta v}{t} \\ \\ :\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time \: taken} \\ \\ :\implies \sf \vec{a} = \dfrac{\vec{v} - \vec{u}}{t} \\ \\ :\implies \sf -10 = \dfrac{0-30}{t} \\ \\ :\implies \sf -10 = \dfrac{-30}{t} \\ \\ :\implies \sf t = \dfrac{-30}{-10} \\ \\ :\implies \sf t = \dfrac{30}{10} \\ \\ :\implies \sf t \: = 3 \: sec \\ \\ :\implies \sf Time \: taken \: = 3 \: seconds

  • Therefore, time taken to reaches the maximum height = 3 sec!

Choice may vary and yours!

✡️ Now let us find out the maximum height by using third equation of motion!

:\implies \sf \vec{v}^{2} - \vec{u}^{2} \: = 2\vec{a}s \\ \\ :\implies \sf (0)^{2} - (30)^{2} = 2(-10)(s) \\ \\ :\implies \sf 0 - 900 = -20s \\ \\ :\implies \sf -900 = -20s \\ \\ :\implies \sf 900 = 20s \\ \\ :\implies \sf \dfrac{900}{20} \: = s \\ \\ :\implies \sf \dfrac{90}{2} \: = s \\ \\ :\implies \sf s \: = 45 \: m \\ \\ :\implies \sf Distance \: = 45 \: m

  • Therefore, distance = 45 m!

✡️ Now by using second equation of motion let us calculate the maximum height!

:\implies \sf s \: = \vec{u}t + \dfrac{1}{2} \: \vec{a}t^2 \\ \\ :\implies \sf s \: = 30(3) + \dfrac{1}{2} \times (-10)(3)^{2} \\ \\ :\implies \sf s \: = 90 + \dfrac{1}{2} \times (-10)(9) \\ \\ :\implies \sf s \: = 90 + \dfrac{1}{2} \times -90 \\ \\ :\implies \sf s \: = 90 + 1 \times -45 \\ \\ :\implies \sf s \: = 90 - 45 \\ \\ :\implies \sf s \: = 45 \: m \\ \\ :\implies \sf Distance \: = 45 \: m

  • Therefore, distance = 45 m!

Choice may vary and yours!


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