A ball is dropped from the top of a 20m high tower. If it's velocity is increasing with an accelerate of 10m/s² then at what speed will the ball hit the ground? How much time will it take to hit the ground?
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A ball is dropped from the top of a 20m high tower. If it's velocity is increasing with an accelerate of 10m/s² then at what speed will the ball hit the ground? How much time will it take to hit the ground?
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Answer:
ᴿᴱᶠᴱᴿ ᵀᴴᴱ ᴬᵀᵀᴬᶜᴴᴹᴱᴺᵀ
ᴴᴼᴾᴱ ᴵᵀ ᴴᴱᴸᴾˢ ◉‿◉
GivEn :
To find :
SoluTion :
We'll solve this question by using the equations of motion.
Eqn used : v² = u² + 2as ; v = u + at
Substituting the values,
[tex]\implies \: \: \sf{v}^{2} - {u}^{2} = 2as \\ \\ \\[/tex]
[tex]\implies \: \: \sf{v}^{2} - {0}^{2} = 2 \times 10 \times 20 \\ \\ \\ [/tex]
[tex]\implies \: \: \sf{v}^{2} = 400 \\ \\ \\ [/tex]
[tex]\implies \: \: \sf v = 20 \: m {s}^{ - 1} \\ \\ \\[/tex]
Now,
Time required by the ball to hit the ground :
[tex] \implies \sf \: \: v = u + at \\ \\ \\ [/tex]
[tex]\implies \sf \: \: 20 = 0+ 10 \times t \\ \\ \\ [/tex]
[tex]\implies \sf \: \: 20 = 10t \\ \\ \\[/tex]
[tex]\implies \sf \: \: t = \cancel\dfrac{20}{10} \\ \\ \\ [/tex]
[tex]\implies \sf \: \: t = 2 \: s \\ \\ \\ [/tex]
Therefore, the speed required to hit the ground is 20 m/s and the time taken to hit the ground is 2 s .