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Question: A particle is projected with a velocity \(6 \hat { \dot { i } } + 8 \hat { \mathrm { j } }\) , 2m a...

A particle is projected with a velocity 6i˙^+8j^6 \hat { \dot { i } } + 8 \hat { \mathrm { j } } , 2m away from a vertical wall. After striking the wall it lands at …………… away from the wall. B

A

3 m

B

3.3 m

C

5.5 m

D

6.6 m

Answer

6.6 m

Explanation

Solution

T = 2uyay=2×810=1.6 s\frac { 2 \mathrm { u } _ { \mathrm { y } } } { \mathrm { a } _ { \mathrm { y } } } = \frac { 2 \times 8 } { 10 } = 1.6 \mathrm {~s}

t = 3/6 = 0.5 s

x = ux (T - t)

= 6(1.6 – 0.5) = 6.6 m