Question

A ball is gently dropped from a height of 20 m. If its velocity
increases uniformly at the rate of 10 m s2, with what velocity
will it strike the ground? After what time will it strike the
ground?​

Answer 1

Answer:

⇒ Initial velocity (u) = 0 m/s

⇒ Final velocity (v) = ? m/s

⇒ Ball falls a total distance (s) = 20 m

⇒ Acceleration (a) = 10 m/s²

⇒ Time taken to strike the ground (t) = ? seconds

We Know That,

$\bigstar$ Equation of motion

$\green{\boxed{\rm v^2 -u^2 = 2as}}$

$\rm \implies v^2-u^2 = 2as$

Substituting the values in the above equation

We get,

$\implies \rm v^2 - 0 = 2 \times 10 \times 20$

$\implies \rm v^2 = 400$

$\implies \rm v = 20 \; m/s$

$\blue{\boxed{\sf v = 20\;m/s}}$

Now to find time taken (t),

Using the formula,

$\bigstar$ Equation of motion

$\orange{\boxed{\rm {v=u+at}}}$

$\implies \rm 20 = 0 +10 \times t$

$\implies \rm 20 = 10t$

$\implies \rm t = 2\;s$

$\red{\boxed{\sf t = 2\;s}}$

Hence,

Final velocity (v) = 20 m/s

Time (t) = 2 seconds

Answer 2

Explanation:

Answer:

⇒ Initial velocity (u) = 0 m/s

⇒ Final velocity (v) = ? m/s

⇒ Ball falls a total distance (s) = 20 m

⇒ Acceleration (a) = 10 m/s²

⇒ Time taken to strike the ground (t) = ? seconds

We Know That,

\bigstar★ Equation of motion

\green{\boxed{\rm v^2 -u^2 = 2as}}

v

2

−u

2

=2as

\rm \implies v^2-u^2 = 2as⟹v

2

−u

2

=2as

Substituting the values in the above equation

We get,

\implies \rm v^2 - 0 = 2 \times 10 \times 20⟹v

2

−0=2×10×20

\implies \rm v^2 = 400⟹v

2

=400

\implies \rm v = 20 \; m/s⟹v=20m/s

\blue{\boxed{\sf v = 20\;m/s}}

v=20m/s

Now to find time taken (t),

Using the formula,

\bigstar★ Equation of motion

\orange{\boxed{\rm {v=u+at}}}

v=u+at

\implies \rm 20 = 0 +10 \times t⟹20=0+10×t

\implies \rm 20 = 10t⟹20=10t

\implies \rm t = 2\;s⟹t=2s

\red{\boxed{\sf t = 2\;s}}

t=2s

Hence,

Final velocity (v) = 20 m/s

Time (t) = 2 seconds