Question

. an object is thrown vertically upward with initial velocity 40 m/s . than what will be the maximum height reached by the object (take g=10m/s2)​

Answer 1

Answer:

Initial Velocity (u) = 40m/s

Final Velocity (v) = 0m/s { after height , h}

g = -10m/s²

Using velocity-position relation:

v² = u² + 2gh

0² = 40² + 2×(-10)×h

-1600 = -20h

h = 80m.

Hope it helps you!

Answer 2

Figure regards this question:

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Provided that:

• Initial velocity = 40 m/s
• g = -10 m/s sq.
• Final velocity = 0 m/s

Don't be confused!

• I write g as -10 m/s sq. except of +10 m/s sq. because the object is thrown upwards that's why acceleration due to gravity, g came as negative.

• Final velocity cames as zero because when the object is thrown then it stops at a particular point or at a height therefore, it's final velocity cames as zero.

To calculate:

• The maximum height

Solution:

• The maximum height = 80 metres

Using concept:

• Third equation of motion

Using formula:

• ${\small{\underline{\boxed{\sf{v^2 \: - u^2 \: = 2gs}}}}}$

Where, v denotes final velocity, u denotes initial velocity, g denotes acceleration due to the gravity and s denotes displacement or distance or height!

Required solution:

$:\implies \sf v^2 \: - u^2 \: = 2gs \\ \\ :\implies \sf (0)^{2} - (40)^{2} = 2(-10)(s) \\ \\ :\implies \sf 0 - 1600 = (-20s) \\ \\ :\implies \sf -1600 = -20s \\ \\ :\implies \sf 1600 = 20s \\ \\ :\implies \sf \dfrac{1600}{20} \: = s \\ \\ :\implies \sf \cancel{\dfrac{1600}{20}} \: = s \: (Cancelling) \\ \\ :\implies \sf 80 \: = s \\ \\ :\implies \sf s \: = 80 \: m \\ \\ :\implies \sf Height \: = 80 \: metres$