A ball is thrown up with a speed of 5 m/s. How high will it go before it begins to fall? [h= 1.27 m]
Answers
Answer:
Given:
Initial speed (u) = 5 m/s
Final speed (v) = 0 m/s
Acceleration due to gravity (g) = -9.8 m/s²
To Find:
Height (h) reached by ball before it begins to fall
Explanation:
Substituting values of v, u & g in the equation:
Height (h) reached by ball before it begins to fall = 1.27 m
Answer:
1.25 m or 1.27 m
Explanation:
Given that, a ball is thrown up with a speed of 5 m/s. (u is 5 m/s, at highest point final velocity i.e. v becomes 0 m/s and a is -10 m/s² as it is against the motion)
We have to find the height or distance.
Using the third equation of motion,
v² - u² = 2as
Substitute the values,
→ (0)² - (5)² = 2(-10)s
→ 0 - 25 = -20s
→ 20s = 25
Divide by 20 on both sides,
→ 20s/20 = 25/20
→ s = 1.25
Hence, the height or distance covered by the ball is 1.25 m.
If we take a is 9.8 them s is 1.27 m.