Question

An object is thrown vertically upwards with a velocity of 20 ms-1. find
(1)the distance travelled by the object to reach the highest point.
(2)the time taken by the object to reach the highest point.
(take g=10m-1)​​

Answer 1

Given

An object is thrown vertically upwards with a velocity of 20 ms-1.

To find

(1)the distance travelled by the object to reach the highest point.

(2)the time taken by the object to reach the highest point.

(take g=10m-1)

Solution

• Initial velocity (u) = 20m/s
• Final velocity (v) = 0
• g = - 10m/s² {thrown vertically upward}
• Distance travelled (h)= ?
• Time taken (t) = ?

→ v² - u² = 2gh

→ (0)² - (20)² = 2*(-10)*h

→ 0 - 400 = -20h

→ -400 = -20h

→ h = 400/20 = 20m

Now,

• t = ?

→ v = u + gt

→ 0 = 20 + (-10)t

→ 0 = 20 -10t

→ 10t = 20

→ t = 20/10 = 2sec

Hence, the distance travelled by object is 20m and time taken to reach the highest point is 2sec

Answer 2

GIVEN:

• Initial Velocity (u) of an object = 20m/s
• Final velocity (v) = 0
• g = -10 m ( Since the object thrown vertically upward )

TO FIND:

• What is the distance travelled by the object to reach the highest point. ?
• What is the time taken by the object to reach the highest point ?

SOLUTION:

❶ We have to find the distance covered by the object to reach the highest point

According to question:

On putting the given values in the equation, we get

$\bf{ v^2 - u^2 = 2gh}$

$\rm{\rightharpoonup (0)^2 - (20)^2 = 2\times -10 \times h}$

$\rm{\rightharpoonup 0 - 400 = -20h}$

$\rm{\rightharpoonup \cancel{-} 400 = \cancel{-}20h}$

$\rm{\rightharpoonup h = \cancel\dfrac{400}{20}}$

$\large\bf{\rightharpoonup \star \: h = 20 \: m \: \star}$

Distance = h = 20 m

❷ Now, we have to find the time taken by the object to cover the distance

According to question:-

On putting the given values in the formula, we get

$\bf{ v = u + gt}$

$\rm{\rightharpoonup 0 = 20 + (-10) \times t}$

$\rm{\rightharpoonup 0 = 20-10 t}$

$\rm{\rightharpoonup 10t = 20}$

$\rm{\rightharpoonup t = \cancel\dfrac{20}{10}}$

$\large\bf{\rightharpoonup \star \: t = 2 \: seconds \: \star}$

Time = t = 2 seconds

❝ Hence, the distance covered by the object to reach the highest point is 20 m and the time taken by the object to reach the highest point is 2 seconds ❞

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