Question

an object thrown vertically upward with an initial velocity of hundred metre per second find A time taken to reach maximum height B maximum height reached C velocity after=5second take g =10 metre per second square​

Answer 1

Answer:

Time taken to reach maximum height = 10 s

Maximum height reached = 500 m

Velocity after 5 seconds = 50 m/s

Explanation:

Given:

• Initial velocity of the object = 100 m/s
• Final velocity of the object = 0 m/s

To Find:

• Time taken to reach maximum height
• Maximum height reached
• Velocity after 5 seconds

Solution:

Time taken to reach maximum height:

By the first equation of motion,

v = u + at

v = final velocity

u = initial velocity

a = acceleration = -g = -10 m/s²

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

Substitute the data,

0 = 100 + -10t

-10t = -100

t = 10 s

Hence time taken to reach the maximum height is 10 s.

Maximum height reached:

By the third equation of motion,

v² - u² = 2as

where v = final velocity

u = initial velocity

a = acceleration = g = -10 m/s²

Substitute the data,

0² - 100² = 2 × -10 × s

-10000 = -20 s

-10000/-20 = s

s = 500 m

Hence the maximum height reached by the object is 500 m.

Velocity after 5 seconds:

By the first equation of motion,

v = u + at

where t = 5 seconds

v = 100 + -10 × 5

v = 100 - 50

v = 50 m/s

Hence the velocity of the object after 5 seconds is 50 m/s.

Answer 2

u = 100

At maximum height v = 0

s = maximum height

acceleration a = - 10

3rd equation of motion

v ^2 = u^2 + 2as

0 = 100^2 + 2(-10)s

s = 10000/20

s = 500 meter

To find time taken (t):

By first equation of motion

v = u + at

0 = 100 + (-10) t

t = 10 s

To find velocity after 5 s (v):

v = u + at

v = 100 + (-10)5

v = 50m/s