Physics, asked by Anonymous, 11 months ago

A ball is thrown vertically upwards with a speed 20m/s. Find
the maximum height reached. When will it reach the
ground? (g =10m/s 2 )

Answers

Answered by sparshu01
40

Explanation:

v = final velocity

u = initial velocity

a = acceleration

t = time

s = distance

In your question, the initial velocity is given as 20m/s , i.e., u=20m/s , the final velocity that the ball can achieve at the maximum height is 0m/s , hence, v=0m/s . Since the only first that cause the acceleration is gravity, a is taken as g where g is acceleration due to gravity, and had a value of 9.81m/s2 . But for simplicity, we can take the value of a to be 10m/s2 , so a=10m/s2 . Now, we need to find, what's s and t.

Note: Since the ball is thrown upwards, which is against the force of gravity (gravity always acts downwards), we need take the value of a (in this case, g) as −10m/s2 .

Using the first equation,

v=u+at

0=20−10t

10t=20

t=2

Using the third equation,

v2=u2+2as

02=202+2×(−10)×s

20s=400

s=20

Hence, the ball will travel for 2 seconds and complete a distance of 20 metres upwards.

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Answered by Anonymous
9

v^{2} - u^{2} = 2gS \\</p><p></p><p>u = 0\\</p><p>v=20m/s\\</p><p>g=10m/s^{2}\\</p><p> v  = \sqrt{2gh}  \\ v ^{2}  = 2gh \\ 20 ^{2}  \div  2 = 10 \: \times h \\ 200 \div 10 = h \\ h = 20m(approx)

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