Question

an object thrown vertically upwards with velocity 20m/s find height it time taken to return the thrower

Answer 1

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instead of object "ball " is taken


The initial velocity is u = 20 m/s

We can use the following equations of motion to solve this question:

v = u + at

s = ut + 1/2 at^2

v^2 = u^2 + 2as

Here, 'u' is the initial velocity, 'v' is the final velocity, 't' is the time taken, 'a' is the acceleration and 's' is the distance traveled. 

In the case of vertical motion (such as this one), acceleration due to gravity is the only acceleration. For simplicity, let us use a value of 10 m/s^2 for it.

a) Maximum height corresponds to a velocity of 0 m/s and after this instant the ball falls down.

Thus, 0^2 = 20^2 + 2(-10)s

Negative acceleration is used to denote the downward direction of acceleration, while the ball is traveling upwards.

Simplifying further, s = 20^2 /(2 x 10) = 20 m.

Thus, the ball will attain a maximum height of 20 m.

b) For the time taken to reach the maximum height:

v = u + at

0 = 20 +(-10) t

So, t = 2 s 

c) Since, the motion of the ball is symmetrical, it will take 2 seconds to rise upwards and 2 seconds to fall back to the ground. Thus, the total journey takes 4 s. (you can also check it by using 20 = 1/2 (10) t^2, for the downward motion of ball).

The velocity at the time of reaching the ground is:

v = u + at

v = 0 + (-10) x 2 = 20 m/s (in the downwards direction).

Hope this helps.

Answer 2

Given,

Velocity of the object=20m/s

To Find,

Height attained by the object.

Solution,

Initial velocity of the object(u)=20m/s

Final velocity (v)=0

We have to find maximum height attained by object

Acceleration due to gravity=-10m/s^2 (opposite to gravity)

Apply equation of motion to find the height attained by it,

V²-u²=2as

⇒-400=-2×10×s

⇒S=20m

Hence, height attained by the object is 20m.