From 18 m height above the ground a ball is dropped from rest

Question

From 18 m height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is ______ m.

(Take $g = 10 {m/s}^{2}$ and neglect the air resistance) [2026]

Smart Achievers · Accepted Answer

(13)

$g = 10 {m/s}^{2}$

$u = 0$

$using v^{2} = u^{2} + 2 a s$

$v^{2} = 2 g (18 - h)$

$Mag. of velocity = mag. of acceleration$

$v = 10 m/s$

$100 = 2 \times 10 (18 - h)$

$\Rightarrow h = 13 m$

Solution

Q.
From 18 m height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is ______ m.

(Take $g = 10 {m/s}^{2}$ and neglect the air resistance) [2026]

Ans.
(13)

$Distance fallen = 18 - h$

$u = 0$

$using v^{2} = u^{2} + 2 a s$

$v^{2} = 2 g (18 - h)$

$Mag. of velocity = mag. of acceleration$

$v = 10 m/s$

$100 = 2 \times 10 (18 - h)$

$\Rightarrow h = 13 m$

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Solution

Q. From 18 m height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is ______ m. (Take g=10 m/s2 and neglect the air resistance) [2026]

Ans. (13) Distance fallen=18-h u=0 using v2=u2+2as v2=2g(18-h) Mag. of velocity=mag. of acceleration v=10 m/s 100=2×10(18-h) ⇒h=13 m

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Q.
From 18 m height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is ______ m.

(Take $g = 10 {m/s}^{2}$ and neglect the air resistance) [2026]

Ans.
(13)

$Distance fallen = 18 - h$

$u = 0$

$using v^{2} = u^{2} + 2 a s$

$v^{2} = 2 g (18 - h)$

$Mag. of velocity = mag. of acceleration$

$v = 10 m/s$

$100 = 2 \times 10 (18 - h)$

$\Rightarrow h = 13 m$