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
So,
v^2-u^2=2as
5^2-0^2=2*10*s
25=20*s
s=25/20
s=5/4
By applying s=ut+1/2at^2,we get,
s=ut+1/2at^2
5/4=0+1/2*10*t^2
5/4*1/5=t^2
t=1/2 sec or 0.5 sec.
Answer:
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