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)
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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.
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