a stone is thrown in a vertically upward direction with a velocity of 5 m s -1. if the acceleration of the stone during its motion is 10 m s–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answers
Answer:
1.25 m
Explanation:
Initial velocity = 5 m/s
g = -10 m/s²
at highest point v = 0 m/s
apply, v = u + at
0 = 5 - 10t
t = 2 s
v² - u² = 2 * a * s
0² - 5² = 2 * -10 * s ⇒ -25 = -20 * displacement ⇒ displacement = 1.25 m
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Answer:
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initial velocity (u)=5 m/s
initial velocity (u)=5 m/sa= -10m/s² ( negative as downward)
initial velocity (u)=5 m/sa= -10m/s² ( negative as downward)final velocity (v) = 0 m/s (as ball comes to rest at highest position)
initial velocity (u)=5 m/sa= -10m/s² ( negative as downward)final velocity (v) = 0 m/s (as ball comes to rest at highest position)Now,
initial velocity (u)=5 m/sa= -10m/s² ( negative as downward)final velocity (v) = 0 m/s (as ball comes to rest at highest position)Now, 2as = v² - u²
initial velocity (u)=5 m/sa= -10m/s² ( negative as downward)final velocity (v) = 0 m/s (as ball comes to rest at highest position)Now, 2as = v² - u² 2x(-10)xs = 0 - 5²
initial velocity (u)=5 m/sa= -10m/s² ( negative as downward)final velocity (v) = 0 m/s (as ball comes to rest at highest position)Now, 2as = v² - u² 2x(-10)xs = 0 - 5² -20s = - 25
initial velocity (u)=5 m/sa= -10m/s² ( negative as downward)final velocity (v) = 0 m/s (as ball comes to rest at highest position)Now, 2as = v² - u² 2x(-10)xs = 0 - 5² -20s = - 25 s = -25/-20 = 5 / 4
initial velocity (u)=5 m/sa= -10m/s² ( negative as downward)final velocity (v) = 0 m/s (as ball comes to rest at highest position)Now, 2as = v² - u² 2x(-10)xs = 0 - 5² -20s = - 25 s = -25/-20 = 5 / 4 = 1.25 m <-- height attained by stone
initial velocity (u)=5 m/sa= -10m/s² ( negative as downward)final velocity (v) = 0 m/s (as ball comes to rest at highest position)Now, 2as = v² - u² 2x(-10)xs = 0 - 5² -20s = - 25 s = -25/-20 = 5 / 4 = 1.25 m <-- height attained by stoneNow , v = u + at
initial velocity (u)=5 m/sa= -10m/s² ( negative as downward)final velocity (v) = 0 m/s (as ball comes to rest at highest position)Now, 2as = v² - u² 2x(-10)xs = 0 - 5² -20s = - 25 s = -25/-20 = 5 / 4 = 1.25 m <-- height attained by stoneNow , v = u + at 0 = 5 + (-10)t
initial velocity (u)=5 m/sa= -10m/s² ( negative as downward)final velocity (v) = 0 m/s (as ball comes to rest at highest position)Now, 2as = v² - u² 2x(-10)xs = 0 - 5² -20s = - 25 s = -25/-20 = 5 / 4 = 1.25 m <-- height attained by stoneNow , v = u + at 0 = 5 + (-10)t t = -5/-10
initial velocity (u)=5 m/sa= -10m/s² ( negative as downward)final velocity (v) = 0 m/s (as ball comes to rest at highest position)Now, 2as = v² - u² 2x(-10)xs = 0 - 5² -20s = - 25 s = -25/-20 = 5 / 4 = 1.25 m <-- height attained by stoneNow , v = u + at 0 = 5 + (-10)t t = -5/-10 = 0.5 s <-- time taken to reach highest point