Physics, asked by JayKumar163, 1 year ago

With what speed must a ball be thrown vertically up in order to rise to a maximum height of 53.7 m?

Answers

Answered by gaurang5

hello friend!

here h= 57.3
v=0
a= -10m/sec^2

by v^2 = u^2 - 2gh
0= u^2 -20 × 57.3
u^2 = 1146
u = √1146
u= 33.8526 m

JayKumar163: But ans is 32.77 m/s and u have to calculate the speed!

gaurang5: oh this is because i took g as 10 you can place 9.8 there and can get the right answer

JayKumar163: But why do we take g as 10??

gaurang5: because gravitational aceelaration is 9.8 but for the sake of ease we take it as 10

JayKumar163: Ohkay and one more question, why u have taken g in (-)?

gaurang5: g in - because we have threw the ball upwards while g acts in downward direction or opposite direction

gaurang5: if we have threw the ball downward g is taken as +

JayKumar163: Thanks. I have marked your answer as brainliest!

gaurang5: thanks a lot

Answered by shirleywashington

Answer:

Initial speed of ball, u = 32.44 m/s

Explanation:

Maximum height, h = 53.7 m

Let the initial speed of the ball is u and at maximum height its final speed becomes zero. We have to find u. It can be calculated using third equation of motion as :

$v^2-u^2=2ah$

Here, a = g = -9.8 m/s²

So, $u^2=2gh$

$u^2=2\times 9.8\ m/s^2\times 53.7\ m$

$u^2=1052.52\ m$

u = 32.44 m/s

Hence, the speed with which the ball is thrown vertically up is 32.44 m/s

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