A Thoughtful Journey

Classical Mechanics Level 3

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 $60$ 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 $m/s^{2}$ , is

Note : Angle is in Degree Measure , g = $10 m/s^{2}$ and take $\sqrt{3} = 1.73$

The answer is 5.

1 solution

Gokul Kumar
Aug 2, 2015

the velocity will be in u sin60 ----> 10 (√3/2)----->5√3 m/s

time taken will be (2 u)/g ------->(2 5√3)/10------>√3 sec

now using the formula s= u*t + 1/2 a t^2

we get , 1.15 = 5√3 +1/2 * a * 3

taking √3=1.73 and solving the equation we get a=5m/s^2 .

Simple Standard Solution , but mention that ' $u$ ' you are taking is the vertical component of velocity when finding time.

Venkatesh K - 5 years, 10 months ago

oh ya ok (y)

Gokul Kumar - 5 years, 10 months ago

Shouldn't acceleration be negative?Shouldn't it be -5 m/s^2?

Jacob Sony - 5 years, 10 months ago

Nope, it's the pseudo acceleration that's negative.

Kishore S. Shenoy - 5 years, 9 months ago

A Thoughtful Journey

The answer is 5.

1 solution

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