Question

An object is thrown vertically upwards and
rises to a height of 10 m. Calculate
the velocity with which the object was thrown
upwards and
the time taken by the object to reach the highest point. (g = 9.8 m/s2)​

Answer 1

Explanation:

s=h=10m

v=0 (final velocity)

a=g=9.8 m/s2

we know that eq. of motion

v2=u2+2as

0=u2+2*9.8*10

-u2=196

-u =+-14

u=1 4

Hence, 14 m/s velocity object through upwards

and

v=u+at

0=14+(-9.8)t

9.8t =14

t=14/9.8

t=1.5 (approx)

Answer 2

Given:-

• Maximum height = H = 10m
• g = 9.8 m/s²

To find:-

• Initial velocity with which the object was thrown = u
• Time of flight = t

Answer:-

Considering the motion from the instant where the ball is thrown till it reaches the maximum height:-

• final velocity = v = 0 m/s

(This is because, if it was not 0, the object could still travel upwards, and hence it could never reach the maximum height. But since hee it is maximum height, that means final velocity = 0 m/s)

So,

v² = u² + 2as

Sign convention: Acceleration will be negative (-g) since it is acting in downward direction, but displacement will be positive as it is in upward direction. Putting available values,

→ 0² = u² + (2 × -9.8 × 10)

→ 2 × 9.8 × 10 = u²

→ u² = 196

→ u = ±14 m/s

But -14 m/s means object was thrown downwards, but since here the object is thrown upwards, we have to consider positive value.

→ u = 14 m/s Ans.

Again,

Considering the motion from the instant where the ball is thrown till it reaches the maximum height:-

v = u + at

Putting available values,

→ 0 = 14 + (-g × t)

→ 9.8 × t = 14

→ t = 14/9.8

→ t = 1.43 seconds Ans.