# 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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by pahiroy1221

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by pahiroy1221

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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...

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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