Question

ball is thrown vertically upward, its velocity becomes zero after 6 sec, Calculate the maximum height reached by the body (g=10 m/s²​

Answer 1

Given:-

• Final velocity of ball = 0m/s

• Time taken = 6s

• Acceleration due to gravity = -9.8m/s²

To Find:-

• Highest Distance reached by ball

Formulae used:-

• v = u + at

• S = ut + ½ × a × t²

Now,

→ v = u + at

→ 0 = u + -10 × 6

→ -u = -60

→ u = 60

Hence, The Initial Velocity of ball is 60m/s

Therefore,

→ s = ut + ½ × a × t²

→ s = 60 × 6 + ½ × -10 × (6)²

→ s = 360 - 5 × 36

→ s = 360 - 180

→ s = 180m

Hence, The highest Distance reached by ball is 180m.

Answer 2

Given:

⠀⠀

• A ball is thrown vertically upwards.
• Its velocity becomes 0 after 6 sec. It means final velocity of ball = 0 m/s
• Accleration due to gravity = 10 m/s$^{2}$ (assuming)

⠀⠀

To find:

⠀⠀

Maximum height reached by the ball.

⠀⠀

Solution:

⠀⠀

Using first equation of motion (i.e. v = u + at) we will firstly find initial velocity of the ball.

⠀⠀

v = u + at

➝ 0 = u + (–10 × 6)

➝ –u = –60

➝ u = 60 m/s

⠀⠀

Now, using second equation of motion

⠀⠀

s = ut + 1/2at$^{2}$

➝ s = 60 × 6 + 1/2 ×(–10) × 6$^{2}$

➝ s = 360 + 1/2 × (–10) × 36

➝ s = 360 – 5 × 36

➝ s = 360 – 180

➝ s = 180

⠀⠀

_______________________________

⠀⠀

Maximum height reached by the body = 180 m.

_____________________________

Some other formulas:

• First equation of motion

v = u + at

• Second equation of motion

s = ut + 1/2at$^{2}$

• Third equation of motion

v$^{2}$ = u$^{2}$ + 2gh