Question

◆ A ball thrown up vertically returns to the thrower after 6 sec :
A) The velocity with which it was thrown.
B) Maximum hieght.
C) Position after 4 sec.

Answer 1

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$a ) \: \: \: \: \: \: \: \: \: \: \: \: \bf\red{➾}\bf{Formula}$

$\bf \red{= >} \bf{v=u+gt}$

$\bf\red{➾} \: 0 = u + ( -10 )( 3 )$

$\bf\red{➾} \: -4 = -30$

$\bf\red{➾} \: u = 30ms {}^{ - 1}$

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$\bf{b)}\bf\red{➾} \: \: v {}^{2} - u {}^{2} = 2gh$

$\bf\red{➾} \: \: 0 - u {}^{2} - u {}^{2} = 2gh$

$\bf\red{➾} \: \: \: h \: = \frac{ - u {}^{2} }{2g}$

$\bf\red{➾} \: h = \frac{ - (30) {}^{2} }{2 \times ( - 10)}$

$\bf\red{➾} \: \frac{ \cancel{ - } \cancel{90} \cancel{0}}{ \cancel{ - } \cancel{2} \cancel{0}}$

$\bf\red{➾} \: h = 45m$

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$\bf{c)} \: \: \: \bf\red{➾} \: s = ut + \frac{1}{2} gt {}^{2}$

$\bf\red{➾} \: s \: = 0 \times t + \frac{1}{ \cancel{2}} \times \cancel{10} \: \: {}^{5} \times 1 {}^{2}$

$\bf\red{➾} \: s = 5m$

➥ Position after 4 sec = 45 - 5 = 40 m

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➻ Hope you understand this concept mate ★

Answer 2

Answer:

C

Step-by step explanation:

because it is thrown up vertically returns to the thrower after 6 second.