Question

an object is thrown vertically upwards with a velocity of 98m/s. calculate the maximum height to which it rises and time taken to reach the max height
(g=9.8/s2)​

Answer 1

Answer :

• The maximum height reached by the object is , h = 490 m.

• Time taken by the object to reach the maximum height , t = 10 s.

Explanation :

Given :

• Initial velocity with which the object is thrown vertically upwards , u = 98 m/s.

• Acceleration due to gravity = -9.8 m/s².

[Since , the ball is thrown vertically upwards i.e, against the gravity , thus the acceleration due to gravity will be taken as negative , g = -9.8 m/s²]

• Final Velocity of the object , v = 0 m/s.

[Final Velocity (v) at the maximum height is taken as zero , v = 0]

To find :

• Maximum height reached by the object , h = ?.

• Time Taken by the object to reach the maximum hieght , t = ?.

Knowledge required :

• Third equation of motion '

⠀⠀⠀⠀⠀⠀⠀⠀⠀v² = u² ± 2gh

Where :

• v = final Velocity
• u = initial Velocity
• g = accelaration due to gravity
• h = height

Now, Deriving the equation for maximum height (h) :

From the third equation of motion,

==> v² = u² - 2gh

==> 0² = u² - 2gh

==> -u² = -2gh

==> u² = 2gh

==> u²/2g = h

⠀⠀⠀⠀⠀⠀⠀⠀⠀∴ h = u²/2g.

Hence the equation for finding the maximum height is u²/2g.

• First equation of motion :

⠀⠀⠀⠀⠀⠀⠀⠀⠀v = u ± gt

Where :

• v = final Velocity
• u = initial velocity
• g = accelaration due to gravity
• t = time taken

Solution :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀To find the maximum height reached by the object :

Using the equation for maximum height and substituting the values in it , we get :

==> h = u²/2g

==> h = 98²/(2 × 9.8)

==> h = 98²/19.6

==> h = 9604/19.6

==> h = 490

∴ h = 490 m.

Hence the maximum height reached by the object is 490 m.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀To find the time taken to reach the maximum height :

By using the first equation of motion and substituting the values in it, we get :

==> v = u ± at

==> v = u - at

==> 0 = 98 - 9.8 × t

==> -98 = -9.8t

==> 98 = 9.8t

==> 98/9.8 = t

==> 980/98 = t

==> 10 = t

∴ t = 10 s.

Hence the time taken by the object to reach the maximum hieght is 10 s.

Therefore,

• The maximum hieght (h) reached by the object with initial velocity (v) of 98 m/s under the Acceleration due to gravity (g) of 9.8 m/s² is 490 m.

• The time taken by the object to reach the maximum hieght is 10 s.