A ball is thrown upward and reaches a height of $64 f e e t$ , its initial velocity should be $(g) = 32 f t / s e c^{2})$

(1) 64 f t / s e c

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(2) 72 f t / s e c

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(2) 32 f t / s e c

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(4) 4096 f t / s e c

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Solution

The correct option is A $(1) 64 f t / s e c$
$H m a x = \frac{u^{2}}{2 g}$
$64 = \frac{u^{2}}{2 \times 32}$
$64^{2} = u^{2}$
$u = 64 f t / s e c$

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A ball is thrown upward and reaches a height of 64feet, its initial velocity should be (g)=32ft/sec2)

The correct option is A (1)64ft/secHmax=u22g64=u22×32642=u2u=64ft/sec

A ball is thrown upward and reaches a height of $64 f e e t$ , its initial velocity should be $(g) = 32 f t / s e c^{2})$

The correct option is A $(1) 64 f t / s e c$
$H m a x = \frac{u^{2}}{2 g}$
$64 = \frac{u^{2}}{2 \times 32}$
$64^{2} = u^{2}$
$u = 64 f t / s e c$