A ball is shot vertically upward from the ground- It loses 2-3 rd of initial speed after rising up for 4 seconds- then total time of upward journey will be

The correct option is D 6 sLet initial speed be V_0 After 4 seconds, speed must be V_03 Using, nbsp;v=u+at V_03=V_0 10(4) 40=2V_03 V_0=40× 32=60 m/s Again ...

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Standard XII

Mathematics

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A ball is sho...

Question

A ball is shot vertically upward from the ground. It loses $\frac{2}{3}$ rd of initial speed after rising up for 4 seconds, then total time of upward journey will be

5 s

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6 s

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7 s

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Solution

The correct option is B $6 s$
Let initial speed be $V_{0}$
After $4 seconds$ , speed must be $\frac{V_{0}}{3}$

Using, $v = u + a t$
$\frac{V_{0}}{3} = V_{0} - 10 (4)$
$40 = \frac{2 V_{0}}{3}$
$V_{0} = \frac{40 \times 3}{2} = 60 m/s$

Again using, $v = u - g t$ , but now $v = 0, u = V_{0}$
$t = \frac{V_{0}}{g} = \frac{60}{10} = 6 s$
Hence total time for upward journey is $6 s$ .

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A ball is shot vertically upward from the ground. It loses 23rd of initial speed after rising up for 4 seconds, then total time of upward journey will be

The correct option is B 6 sLet initial speed be V0 After 4 seconds, speed must be V03 Using, v=u+at V03=V0−10(4) 40=2V03 V0=40×32=60 m/s Again using, v=u−gt, but now v=0, u=V0 t=V0g=6010=6 s Hence total time for upward journey is 6 s.

A ball is shot vertically upward from the ground. It loses $\frac{2}{3}$ rd of initial speed after rising up for 4 seconds, then total time of upward journey will be