if a ball is thrown upwards at a speed of 11 m/s from the balcony, 4m above the ground, how much time would it take to strike the ground at the base of the balcony.please answer it quickly.
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Answered by
16
u=11m/s
v=0m/s
a=9.8
t=?
using v=u+at we can find the time taken to get at the highest point in upward motion.
0=11+9.8t
t=11/9.8=1.12 sec
now the total distance to be covered=121/9.6+4=1.17m
so using s=ut+1/2at^2 we can find time taken to reach the ground from the highest point.
t^2=2.07
t=1.44
therefore total time taken to reach the ground =1.44+1.12=2.56 sec
i hope this helps u...
v=0m/s
a=9.8
t=?
using v=u+at we can find the time taken to get at the highest point in upward motion.
0=11+9.8t
t=11/9.8=1.12 sec
now the total distance to be covered=121/9.6+4=1.17m
so using s=ut+1/2at^2 we can find time taken to reach the ground from the highest point.
t^2=2.07
t=1.44
therefore total time taken to reach the ground =1.44+1.12=2.56 sec
i hope this helps u...
rileybiers:
thank u so much
Answered by
8
Taking upward as positive direction,
u = +11 m/s
g = -9.8 m/s²
height of balcony = 4m
when ball reaches ground, displacement, s = -4m
t = ?
s = ut + 1/2 gt²
⇒ -4 = 11t + 1/2 × (-9.8)×t²
⇒ -4 = 11t - 4.9t²
⇒ 4.9t² - 11t - 4 = 0
Solving the quadratic equation, we get
t = 2.56 s and -0.31s
Since time is positive, the time taken to reach ground is 2.56s
u = +11 m/s
g = -9.8 m/s²
height of balcony = 4m
when ball reaches ground, displacement, s = -4m
t = ?
s = ut + 1/2 gt²
⇒ -4 = 11t + 1/2 × (-9.8)×t²
⇒ -4 = 11t - 4.9t²
⇒ 4.9t² - 11t - 4 = 0
Solving the quadratic equation, we get
t = 2.56 s and -0.31s
Since time is positive, the time taken to reach ground is 2.56s
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