A ball is thrown vertically upward and it reaches a height of 90 m . F

1.	A ball is thrown vertically upward and it reaches a height of 90 m . Find (a) the velocity with which it was thrown and (b) the height reached by the ball 7 seconds after it was thrown (g=9.8 ms-2)
Answer» Given:- Final velocity ,V = 0m/s Height ,h = 90m Acceleration due to GRAVITY ,g = 9.8m/s² Time taken ,t = 7s To Find:- (a) Initial velocity ,u (b) height attained by ball in 7 second. Solution:- By using 3rd EQUATION of motion → v² = u² +2as Substitute the value we get → 0² = u² + 2×(-9.8) ×90 → 0 = u² + (-1764) → -u² = -1764 → u² = 1762 → u = √1764 → u = 42m/s ∴ The initial velocity of the ball is 42m/s . Now we have to calculating the height attained by ball in 7 s. Using 2nd equation of motion → h = ut +1/2at² Substitute the value we get → h = 42×7 + 1/2×(-9.8)×7² → h = 294 + (-4.9) ×49 → h = 294 - 240.1 → h = 53.9 m ∴ The height attained by the ball aftern 7 second is 53.9 metre.

A ball is thrown vertically upward and it reaches a height of 90 m . Find (a) the velocity with which it was thrown and (b) the height reached by the ball 7 seconds after it was thrown (g=9.8 ms-2)

Answer»

Given:-

By using 3rd EQUATION of motion

→ v² = u² +2as

Substitute the value we get

→ 0² = u² + 2×(-9.8) ×90

→ 0 = u² + (-1764)

→ -u² = -1764

→ u² = 1762

→ u = √1764

→ u = 42m/s

∴ The initial velocity of the ball is 42m/s .

Now we have to calculating the height attained by ball in 7 s.

Using 2nd equation of motion

→ h = ut +1/2at²

Substitute the value we get

→ h = 42×7 + 1/2×(-9.8)×7²

→ h = 294 + (-4.9) ×49

→ h = 294 - 240.1

→ h = 53.9 m

∴ The height attained by the ball aftern 7 second is 53.9 metre.