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The train runs in a straight line with a constant acceleration "a". A girl standing in a train throws a ball forward at a speed of $10\ \frac{m}{s}$ at an angle of $600$ degrees from the horizontal. The girl must advance $1.15\ meters$ inside the train to catch the ball at the initial height. The acceleration of the train, in
$\frac{m}{s^2},$ is:
Option: 1
$5\ \frac{m}{s^2}$

Option: 2
$10\ \frac{m}{s^2}$

Option: 3
$20\ \frac{m}{s^2}$

Option: 4
$25\ \frac{m}{s^2}$

Answers (1)

best_answer

$R=\frac{1}{2}at^2+1.15$

$u^2\frac{sin2\theta}{g}=\frac{1}{2}a\left(2sin\frac{\theta}{g}\right)^2+\ 1.15$

Or, $\frac{100\times\frac{\sqrt3}{2}}{10}=\frac{1}{2}a\left(\frac{4\times100\times\frac{3}{4}}{100}\right)+1.15$

Or, $2\times5\sqrt3=3a+2.30$

Or, $10\sqrt3=3a+2.30$

Or, $10\times1.73-2.3\ =3a$

Or, $17.3\ -2.3\ =3a$

Or, $a= 5\ \frac{m}{s^2}$

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