Question

log8 to the base 10+log25 to the base 10+2 log3 to the base 10-log18 to the base 10
evaluate this​

Answer 1

Answer:

Given:-

Initial velocity ,u = 5m/s

Final velocity ,v = 0m/s

Acceleration due to gravity ,g= 10m/s²

To Find:-

Maximum height attained by stone ,h

Total time taken to reach maximum height ,t

Solution:-

⠀⠀⠀⠀⠀According to the Question

It is given that the stone is thrown vertically upward direction . So the acceleration due to gravity on stone is negative.

→ g = -10m/s²

Now, firstly we calculate the maximum height attained by the stone . Using 3rd equation of motion

v² = u²+ 2gh

where,

v denote final velocity

u denote initial velocity

g denote acceleration due to gravity

h denote maximum height attained by stone

Substitute the value we get

:\implies:⟹ 0² = 5² + 2(-10) × h

:\implies:⟹ 0 = 25 -20h

:\implies:⟹ -25 = -20h

:\implies:⟹ 25 = 20h

:\implies:⟹ 25/20 = h

:\implies:⟹ h = 25/2

:\implies:⟹ h = 1.25 m

Hence, the maximum height attained by stone is 1.25 metres.

Now , calculating the time taken to reach maximum height .

Using 1st equation of motion

v = u + gt

Substitute the value we get

:\implies:⟹ 0 = 5 + (-10) × t

:\implies:⟹ -5 = -10×t

:\implies:⟹ 5 = 10 × t

:\implies:⟹ 5/10 = t

:\implies:⟹ t = 5/10

:\implies:⟹ t = 0.5s

Hence, the time taken to reach maximum height is 0.5 second .

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