A ball thrown vertically upward reached a maximum height of 20m (i) W

1.	A ball thrown vertically upward reached a maximum height of 20m (i) What was the velocity of the stone at the instant of throwing up? (ii) How much time did the ball take to reach the height 20 m?
Answer» (i) v² = u² + 2as u²= v² - 2as = 0² - 2 x 10 x 20 = 400 u = 20 m/s (ii) v = u + t t = \(\frac{V-U}{a}\) = \(\frac{0-20}{-10}\) = 2 s \(v^2=u^2+2as\) \(u^2=v^2-2as\) \(=0^2-2(-10)(20)\) =400 (i)u=20 m/s \(v=u+at\) 0=20-10t 20=10t (ii) t=2 sec

A ball thrown vertically upward reached a maximum height of 20m (i) What was the velocity of the stone at the instant of throwing up? (ii) How much time did the ball take to reach the height 20 m?

Answer»

(i) v² = u² + 2as

u²= v² - 2as

= 0² - 2 x 10 x 20

= 400

u = 20 m/s

(ii) v = u + t

t = \(\frac{V-U}{a}\)

= \(\frac{0-20}{-10}\)

= 2 s

\(v^2=u^2+2as\)
\(u^2=v^2-2as\)
\(=0^2-2(-10)(20)\)
=400
(i)u=20 m/s

\(v=u+at\)

0=20-10t

20=10t

(ii) t=2 sec

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A ball thrown vertically upward reached a maximum height of 20m (i) What was the velocity of the stone at the instant of throwing up? (ii) How much time did the ball take to reach the height 20 m?

\(v^2=u^2+2as\)\(u^2=v^2-2as\)\(=0^2-2(-10)(20)\)=400(i)u=20 m/s

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\(v^2=u^2+2as\)
\(u^2=v^2-2as\)
\(=0^2-2(-10)(20)\)
=400
(i)u=20 m/s