Question

An object is thrown in vertically
upward direction with a velocity of
10 m/s. calculate th maximum
height attained by the object. Also
find the
time taken to reach?
(take a=10m/s^2).​

Answer 1

Given:-

• Initial velocity,u = 10 m/s

• Acceleration, due to gravity,g = -10 m/s² [ An object goes up ]

• Final velocity,v = 0 m/s

To be calculated:-

Calculate the height,h and time taken,t ?

Formula used:-

• v² = u² + 2gh

• v = u + gt

Solution:-

( 1 ) From the third equation of motion for freely falling body, we have

v² = u² + 2gh

★ Putting the values in the formula,we get:

⇒ ( 0 )² = ( 10 ) ² × 2 × ( -10 ) × h

⇒ 0 = 100 - 20h

⇒ 20h = 100

⇒ h = 100/20

⇒ h = 5 m

Thus,the maximum height attend by the object is 5 m.

( 2 ) From the first equation of motion for freely falling body, we have

v = u + gt

★ Putting the value in the above formula,we get:

⇒ 0 = 10 + ( - 10 ) × t

⇒ 0 = 10 - 10 t

⇒ 10 t = 10

⇒ t = 10/10

⇒ t = 1 Sec

Thus,the body takes 1 second to reach the highest point.

Answer 2

Heya!!

____________________________________

Given,

Initial velocity = 10m/s

Acceleration = $-10m/s^{2}$

To find:-

➡️ Height

➡️Time taken

Calculation:-

${v}^{2} = {u}^{2} + 2gh$

V = final velocity

u = Initial velocity

g = acceleration

h = height

${0}^{2} = { - 10}^{2} + 2 \times ( - 10) \times h$

$0 = 100 - 20h$

$20h = 100$

$h = \frac{100}{20}$

$\pink{\boxed{h=5m}}$

V = u +gt

V = Final velocity

u = Initial velocity

g = acceleration

t = time

$0 = 10 + ( - 10) \times t$

$0 = 10 - 10t$

$10t = 10$

$t = \frac{10}{10}$

$t = 1$

so,

h = 5m

t = 1sec