Question

A body is thrown up with a velocity 50m/s calculate the maximum height reached and the time taken reach the maximum height a=-10m/s2

Answer 1

$\huge{\boxed{\sf QuEsTIoN}}$

A body is thrown up with a velocity 50m/s calculate the maximum height reached and the time taken reach the maximum height a=-10m/s2

$\huge{\boxed{\sf Answer}}$

t = 5seconds

$\huge{\boxed{\sf Formulas}}$

• v² - u² = 2as
• v = u + at
• s = ut + 1/2 at²

$\huge{\boxed{\sf Solution:-}}$

u = 50m/s ; g = -10m/s²

The velocity at the heighest point will be zero

Thus v = 0

Height can be calculated as

➭ v² - u² = 2gs

➭ 0 -2500 = -20s

➭ s = 2500/20

➭ s = 125m

Time can be calculated as

➭ v = u + gt

➭ 0 = 50 + (-10)t

➭ t = -50/-10

➭ t = 5 sec

☯ Hence Verified

Answer 2

CORRECT QUESTION :-

★ A body is thrown up with a velocity of 50 m/s. Calculate the maximum height reached and the time taken to reach the maximum height { g = -10 m/s². }

GIVEN :-

• Initial velocity , u = 50 m/s.
• final velocity , v = 0.
• acceleration due to gravity , g = -10 m/s².

TO FIND :-

• maximum height , s.
• time taken , t.

SOLUTION :-

★ third equation of motion ★

➳ v² - u² = 2gs.

• u = Initial velocity.
• v = final velocity.
• g = acceleration due to gravity.
• s = maximum height.

➳ (0)² - (50)² = 2(-10)s

➳ 0 - 2500 = -20 s

➳ -2500 = -20 s

➳ s = -2500/-20

➳ s = 2500/20

➳ s = 125 m.

★ first equation of motion ★

✭ v = u + gt.

• u = Initial velocity.
• v = final velocity.
• g = acceleration due to gravity.
• t = time taken.

➝ 0 = 50 + (-10) t

➝ 0 = 50 - 10 t

➝ 0 - 50 = -10 t

➝ -50 = -10 t

➝ t = -50/-10

➝ t = 50/10

➝ t = 5 sec.

Hence the maximum height is 125 m and and time taken to reach the maximum height is 5 sec.