A ball is thrown up with a speed of 15 M per second how high will it go before it begin to falls. (g is equal to 9.8)
Answers
Answered by
48
u = 15 m/s
v = 0 m/s
a = - 9.8 m/s²
s = ?
v² = u² + 2as
= 0 = 225 - 9.8 * 2 * s
= 225 = 9.8*2*s
= s = 225 * 10 / 196
= 11.47 m
Saatwiik:
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Answered by
44
Given
u = 15 m/s
v = 0 (Since at max height, final velocity becomes 0)
a = g = - 9.8 m/s² (negative because the acceleration is downward but the motion is upwards)
Now we know that,
v² - u² = 2gh
=> 0² - 15² = 2 × - 9.8 × h
=> -225 = - 19.6 h
=> h = -225/-19.6
=> h = 225/19.6
=> h = 11.47 m (approx)
Your answer :- 11.47 m
★ To find max height,
v² - u² = 2gh
at v = 0 and when g is negative
=> -u² = - 2gh
=> h = u²/2g
This is the derived formula.
u = 15 m/s
v = 0 (Since at max height, final velocity becomes 0)
a = g = - 9.8 m/s² (negative because the acceleration is downward but the motion is upwards)
Now we know that,
v² - u² = 2gh
=> 0² - 15² = 2 × - 9.8 × h
=> -225 = - 19.6 h
=> h = -225/-19.6
=> h = 225/19.6
=> h = 11.47 m (approx)
Your answer :- 11.47 m
★ To find max height,
v² - u² = 2gh
at v = 0 and when g is negative
=> -u² = - 2gh
=> h = u²/2g
This is the derived formula.
:-)
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