Question

A ball thrown vertically upwards with a speed of 19.6 ms-1 from the top of a tower returns to the earth in 6 seconds. Find the height of the tower. (g = 9.8 ms-2)​

Answer 1

${\large{\pmb{\sf{\underline{RequirEd \: solution...}}}}}$

${\bigstar \:{\pmb{\sf{\underline{Understanding \: the \: question...}}}}}$

⋆ This question says that there is a ball which is thrown upwards in a vertical motion with a speed of ${\sf{(19.6 \: ms^{-1}}}$ It is thrown from the top of a tower and afterthat it returns to the earth surface within 6 seconds. Now this question says that we have to find out the height of the tower. Also this question says that we have to use the value of ${\sf{g \: as \: 9.8 \: ms^{-2}}}$

${\bigstar \:{\pmb{\sf{\underline{Provided \: that...}}}}}$

⠀⠀⠀⠀⠀⠀A ball thrown vertically upwards with a speed of 19.6 ms-1 from the top of a tower returns to the earth in 6 seconds. Find the height of the tower. ${\sf{(g \: = 9.8 \: ms^{-2})}}$

$\sf According \: to \: statement \begin{cases} & \sf{Time \: taken \: = \bf{6 \: seconds}} \\ \\ & \sf{Initial \: velocity \: = \bf{19.6 \: ms^{-1}}} \\ \\ & \sf{Final \: velocity \: = \bf{0 \: ms^{-1}}} \\ \\ & \sf{g \: = \bf{9.8 \: ms^{-2}}}\end{cases}$

Don't be confused! Final velocity came as zero because the ball is at highest point.

⠀⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━

${\bigstar \:{\pmb{\sf{\underline{Using \: concept...}}}}}$

⋆ First equation of motion

⋆ Second equation of motion

${\bigstar \:{\pmb{\sf{\underline{Using \: formulas...}}}}}$

⋆ First equation of motion is given by

⠀⠀⠀⠀⠀⠀${\small{\underline{\boxed{\sf{v \: = u \: + at}}}}}$

⋆ Second equation of motion is given by

⠀⠀${\small{\underline{\boxed{\sf{s \: = ut \: + \dfrac{1}{2} \times a \times t^{2}}}}}}$

(Where, v denotes final velocity , u denotes initial velocity , a denotes acceleration due to gravity , t denotes time and s denotes displacement or distance)

${\bigstar \:{\pmb{\sf{\underline{Full \; Solution...}}}}}$

~ To solve this question firstly we have to use first equation of motion and have to find one more time.

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 0 = 19.6 + (9.8)(t) \\ \\ :\implies \sf 0 = 19.6 + 9.8t \\ \\ :\implies \sf 0 - 19.6 \: = 9.8t \\ \\ :\implies \sf -19.6 \: = 9.8t \\ \\ :\implies \sf \dfrac{-19.6}{9.8} \: = t \\ \\ :\implies \sf -2 \: = t \\ \\ :\implies \sf t \: = 2 \quad \quad \sf (Can't \: be \: negative)\\ \\ :\implies \sf Time \: = 2 \: seconds$

~ Now finding original time by subtracting the new we get or that is already provided.

$:\implies \sf Time \: = 6 - 4 \\ \\ :\implies \sf Time \: = 2 \\ \\ :\implies \sf Time \: = 2 \: seconds$

~ Now we have to use equation two of motion to carry on.

$:\implies \sf s \: = ut \: + \dfrac{1}{2} \times a \times t^{2} \\ \\ :\implies \sf s \: = 0(4) + \dfrac{1}{2} \times 9.8 \times (4)^{2} \\ \\ :\implies \sf s \: = 0 + \dfrac{1}{2} \times 9.8 \times (4)^{2} \\ \\ :\implies \sf s \: = 0 + \dfrac{1}{2} \times 9.8 \times 16 \\ \\ :\implies \sf s \: = 0 + \dfrac{1}{\cancel{{2}}} \times 9.8 \times \cancel{16} \\ \\ :\implies \sf s \: = 0 + 1 \times 9.8 \times 8 \\ \\ :\implies \sf s \: = 0 + 1 \times 78.4 \\ \\ :\implies \sf s \: = 1 \times 78.4 \\ \\ :\implies \sf s \: = 78.4 \\ \\ :\implies \sf Distance \: = 78.4 \: metres$

~ Now at last let's find height of the tower we just have to subtract distance with initial velocity of ball.

$:\implies \sf Height \: of \: tower \: = 78.4 - 19.6 \\ \\ :\implies \sf Height \: of \: tower \: = 58.8 \\ \\ :\implies \sf Height \: of \: tower \: \approx 59 \: metres$

Therefore, 59 m is height of tower.