Question

Stone is vertically thrown upward with an initial velocity 55m/s. Find
a) the maximum height reached by the stone. b) Total time it takes to reach the ground.

Answer 1

Answer:

Max Ht=151.25m and time = 11s

Explanation:

For max height:

at max height body stops momentarily and hence v=0

so

v^2=u^2-2gh => 0=(55)^2-2(10)(h) => h= (55×55)/20 => 151.25=h

Now for time:

Total time of flight for a body projected vertically upward = 2u/g => 2×55/10 => 110/10 => 11s

hope it helped

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Answer 2

$\checkmark$ Given:

Stone is vertically thrown upward with an initial velocity 55 m/s

$\checkmark$ To Find:

a) The maximum height reached by the stone

b) Total time it takes to reach the ground

$\checkmark$ Solution:

We know that,

$\red{\bigstar \ \boxed{\sf h=\dfrac{u^{2}}{2g}}}$

Here,

• h = height
• u = initial velocity
• g = acceleration due to gravity

According to the question,

We need to find,

a) The maximum height reached by the stone

Given that,

Stone is vertically thrown upward with an initial velocity 55 m/s

Hence,

• u = 55 m/s

We know that,

• g = 9.8 m/s²

Substituting the values,

We get,

$\blue{\sf \implies h=\dfrac{(55)^{2}}{(2 \times 9.8)}}$

On further simplification,

We get,

$\green{\sf \implies h=154.33 \ m}$

Therefore,

a) The maximum height reached by the stone = 154.33 m

We know that,

$\red{\bigstar \ \boxed{\sf t=\dfrac{2u}{g}}}$

Here,

• t = time taken
• u = initial velocity
• g = acceleration due to gravity

According to the question,

We need to find,

b) Total time it takes to reach the ground

Given that,

Stone is vertically thrown upward with an initial velocity 55 m/s

Hence,

• u = 55 m/s

We know that,

• g = 9.8 m/s²

Substituting the values,

We get,

$\orange{\sf \implies t=\dfrac{2 \times 55}{9.8} \ s}$

$\pink{\sf \implies t=11.2 \ s}$

Therefore,

b) Total time it takes to reach the ground = 11.2 s