A ball is thrown vertically upward from ground with a speed of 24 5 ms. After what time intervals, the ball vill
be at a height of 294 m from the ground ?
Answers
Correct Question
A ball is thrown vertically upward from ground with a speed of 24.5 m/s. After what time intervals, the ball will be at a height of 29.4 m from the ground ?
Solution-
A ball is thrown vertically upward from ground with a speed of 24.5 m/s.
We have to find that at what time intervals, the ball will be at a height of 29.4 m from the ground.
First equation of motion:
v = u + at
Second equation of motion:
s = ut + 1/2 at²
Third equation of motion:
v² - u² = 2as
From above data we have s = 29.4 m, u = 24.5 m/s and a is -9.8 m/s² (as it is against the motion). So, using the second equation of motion i.e. s = ut + 1/2 at²
Substitute the known values in the above formula,
→ 29.4 = 24.5(t) + 1/2 (-9.8)t²
→ 29.4 = 24.5t - 4.9t²
→ 4.9t² - 24.5t + 29.4 = 0
→ t² - 5t + 6 = 0
→ t² - 3t - 2t + 6 = 0
→ t(t - 3) -2(t - 3) = 0
→ (t - 2)(t - 3) = 0
→ t = 2, 3 sec
Therefore, the time taken by the ball is 2 or 3 sec.
A ball is thrown vertically upward from ground with a speed of 24 5 ms. After what time intervals, the ball will be at a height of 294 m from the ground ?
_______________
✒Given : -
- Speed = 245ms
- Height = 294
✒To Find : -
- Time = ?
✒Formula Used : -
✒On Substituting Values : -
Hence , Time taken by ball is 2 or 3 seconds ..