Question

a stone is thrown vertically upward with a velocity of 40 m/s find the maximum height reached by the stone and time taken to reach the maximum height

Answer 1

Answer:

Maximum height = 81.63 m

Time taken = 4.08 s

Explanation:

Given:

• Initial velocity of the stone = 40 m/s
• Final velocity of the stone = 0 m/s

To Find:

• Maximum height reached
• Time taken to reach the maximum height

Solution:

Maximum height reached:

By the third equation of motion we know that,

v² - u² = 2 as

where v = final velocity

u = initial velocity

a = acceleration

s = distance travelled

Here a = -g ( acceleration due to gravity) = -9.8 m/s²

Here acceleration is taken as negative since the motion of the object is in the direction opposite to that of acceleration due to gravity.

Substituting the data,

0² - 40² = 2 × -9.8 × s

-19.6 s = -1600

s = -1600/-19.6

s = 81.63  m

Hence the maximum height reached by the stone is 81.63 m.

Time taken:

By the first equation of motion,

v = u + at

where v = final velocity

u = initial velocity

a = acceleration = -g

t = time taken

Substitute the data,

0 = 40 + -9.8 × t

-9.8 t = -40

t = -40/-9.8

t = 4.08 s

Hence the time taken to reach the maximum height is 4.08 s.

Answer 2

☆Answer:

●Given:

Initial velocity(u)=40m/s

Final velocity(v)=0m/s

●To prove:

Height of the object(s).

●Proof:

By third equation of motion,

v²-u²=2gs

0-40²=(-2×10×s)

⇒s=160/20=80m/s

Total distance travelled by the stone=upward distance+downward distance=2×s=160m.