An object is thrown vertically upwards with an initial velocity of m/s. (take g=10m/s2 Find –(innotebook)-(i) time taken to reach maximum height(ii) maximum height reached(iii) velocity after 5 seconds

To find the time taken to reach the maximum height, we can use the equation of motion for vertical motion:

v = u + gt

where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.

In this case, the initial velocity is given as m/s and the acceleration due to gravity is 10 m/s^2. Since the object is thrown vertically upwards, the final velocity at the maximum height will be 0 m/s.

So, we can rewrite the equation as:

0 = m/s + (10 m/s^2)t

Solving for t, we get:

t = -u / g

Substituting the given values, we have:

t = -m/s / 10 m/s^2

Simplifying, we find:

t = -0.1 s

Since time cannot be negative, we take the absolute value of t:

t = 0.1 s

Therefore, the time taken to reach the maximum height is 0.1 seconds.

To find the maximum height reached, we can use the equation:

s = ut + (1/2)gt^2

where s is the displacement, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.

At the maximum height, the final velocity is 0 m/s, so we can rewrite the equation as:

0 = m/s * 0.1 s + (1/2)(10 m/s^2)(0.1 s)^2

Simplifying, we find:

0 = 0.1 m + 0.05 m

Therefore, the maximum height reached is 0.15 meters.

To find the velocity after 5 seconds, we can use the equation:

v = u + gt

Substituting the given values, we have:

v = m/s + (10 m/s^2)(5 s)

Simplifying, we find:

v = m/s + 50 m/s

Therefore, the velocity after 5 seconds is m/s + 50 m/s.

To find the time taken to reach the maximum height, we can use the equation of motion for vertical motion:

v = u + gt

where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.

In this case, the initial velocity is given as m/s and the acceleration due to gravity is 10 m/s^2. Since the object is thrown vertically upwards, the final velocity at the maximum height will be 0 m/s.

So, we can rewrite the equation as:

0 = m/s + (10 m/s^2)t

Solving for t, we get:

t = -u / g

Substituting the given values, we have:

t = -m/s / 10 m/s^2

Simplifying, we find:

t = -0.1 s

Since time cannot be negative, we take the absolute value of t:

t = 0.1 s

Therefore, the time taken to reach the maximum height is 0.1 seconds.

To find the maximum height reached, we can use the equation:

s = ut + (1/2)gt^2

where s is the displacement, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.

At the maximum height, the final velocity is 0 m/s, so we can rewrite the equation as:

0 = m/s * 0.1 s + (1/2)(10 m/s^2)(0.1 s)^2

Simplifying, we find:

0 = 0.1 m + 0.05 m

Therefore, the maximum height reached is 0.15 meters.

To find the velocity after 5 seconds, we can use the equation:

v = u + gt

Substituting the given values, we have:

v = m/s + (10 m/s^2)(5 s)

Simplifying, we find:

v = m/s + 50 m/s

Therefore, the velocity after 5 seconds is m/s + 50 m/s.