Question

A train is moving along a straight line with a constant acceleration 'a'. A boy standing in the train throws a ball forward with a speed of $10$ m/s, at an angle of ${60}^o$ to the horizontal. The boy has to move forward by $1.15$m inside the train to catch the ball back at the initial height. The acceleration of the train, in $m/s$, is.(The answer is a single digit integer ranges from 0 to 9)

Answer 1

$uy=10sin60=53\sqrt{m/s}$

$=t=2uyg=2\times 53\sqrt{10}=3\sqrt{s}$

$Sx=uxt+12axt2$

$1.15=5\times t-12a\times t21.15=5\times 3\sqrt{-32a}$

$3a2=5\times 1.73-1.15=8.65-1.15$

$3a2=7.5$

$=a=153=5m/s^2$