Question

6. A stone is thrown vertically upward with a velocity of 30 m/s. How high will it rise? After how
much time will it return to ground?
Ans. 45 m, 6s​

Answer 1

Answer:-

$\red{\bigstar}$ Height

$\large\leadsto\boxed{\sf \blue{45 \: m}}$

$\red{\bigstar}$ Time taken to return to ground

$\large\leadsto\boxed{\sf \blue{6 \: sec.}}$

• Given:-

Stone is thrown vertically upwards with velocity [u] = 30 m/s

• To Find:-

1. Height to which it will rise

2. Time taken to return to ground

• Solution:-

Case 1:

• Using 3rd Equation of motion:-

$\pink{\bigstar}$$\large\boxed{\sf\red{v^2 - u^ 2 = 2gh}}$

here,

v = 0

u = 30 m/s

g = - 10 m/s² [as stone is thrown vertically upwards i.e opposite to gravity]

h = ?

Now,

$\boxed{\bf{v^2 - u^2 = 2 (-g) h}}$

→ (0)² - (30)² = 2 (-10) h

→ 0 - 900 = -20 h

→ h = $\sf\dfrac{-900}{-20}$

→ $\large{\sf\green{h = 45 \: m}}$

Case 2:

• Using 1st Equation of motion:-

$\pink{\bigstar}$$\large\boxed{\sf\red{v = u + gt}}$

→ $\boxed{\bf{v = u + (-g)t}}$

→ 0 = 30 + (-10)t

→ 0 = 30 - 10 t

→ -30 = -10 t

→ t = $\sf\dfrac{-30}{-10}$

→ t = 3 sec.

Therefore,time taken to reach the maximum height will be 3 sec.

• Time taken to return to the ground will be 2 × 3 sec.

= 6 sec.

→ $\large{\sf\green{t = 6 \: sec.}}$

Answer 2

ǫᴜᴇsᴛɪᴏɴ :-

• A stone is thrown vertically upward with a velocity of 30 m/s. How high will it rise?After how much time will it return to ground?

ɢɪᴠᴇɴ :-

• A stone is thrown vertically upward with a :-

• Initial velocity ,u = 30 m/s.

ᴛᴏ ғɪɴᴅ :-

• How high it will rise.

• Time taken for it to reach the ground.

sᴏʟᴜᴛɪᴏɴ :-

Using 3rd equation of motion :-

${\bf\large\blue{→ {v}^{2} - {u}^{2} = 2( - g) \times h }}$

( g is taken as negative because it is acting against the direction of the stone )

${\bf\large\blue{→ 0 - (30)^{2} = 2 \times ( - 10) \times h}}$

${\bf\blue{→ h = \frac{900}{20} }}$

${\bf\blue{→ h = 45 \: m}}$

Using 1st equation of motion :-

${\bf\large\purple{ → v = u + ( - g)t }}$

${\bf\large\purple{→ 0 = 30 - 10t }}$

${\bf\large\purple{→ t = \frac{30}{10} = 3 \: sec }}$

• 3 seconds is the time required to reach the maximum hights.

• Total time to return to the ground,

${\bf\large\pink{ ➯ 2 \times 3 = 6 \: seconds}}$