Question

A stone is thrown upward at 20m/s . The stone reach the ground again .
1). time taken by stone to reach its max height .
2). The height attaind by the height​

Answer 1

Answer:

Let u be the initial velocity and h be the maximum height attained by the stone.

v

1

2

=u

2

−2gh,

(10)

2

=u

2

−2×10×

2

h

100=u

2

−10h ....(i)

Again at height h,

v

2

2

=u

2

−2gh

(0)

2

=u

2

−2×10×h

u

2

=20h ... (ii)

So, from Eqs. (i) and (ii) we have

100=10h

h=10m

Answer 2

Given

• A stone is thrown upward
• Initial Velocity = 20 m/s
• The stone comes back to the ground again

To Find

• Time taken by the stone to reach its maximum height
• The Height attained by the stone

Solution

● We shall find the acceleration with the help of the first equation of motion and the distance with the help of the third equation of motion

✭ Time taken by the stone :

→ v = u+at

• v = Final Velocity = 0 m/s
• u = Initial Velocity = 20 m/s
• a = Acceleration = -10 m/s²
• t = time = ?

→ 0 = 20 + (-10) × t

→ 0-20 = (-10)t

→ -20/-10 = t

→ Time = 10 sec

━━━━━━━━━━━━━━

✭ Maximum Height :

→ v²-u² = 2as

→ 0²-20² = 2×(-10)×s

→ -400 = -20s

→ -400/-20 = s

→ Distance = 20 m