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# A person is standing on a truck moving with a constant velocity 15 m/s. The man throws a ball in such a way that it returns to his hand after the truck has moved 60 m. Find speed and angle of projection: a) as seen from truck b) as seen from the road. Please show the steps.

Ashwin Maran
18 Points
7 years ago
Since the ball returns to the hand after traveling for 60m at a speed 15m/s, time period, T = 60/15 = 4s. As seen from the truck, the ball is obviously thrown at angle of 90, as it cannot reach the thrower`s hand, if it has any x velocity. Using first law of motion, 2u = gt. So, u = 2g m/s. u will be the same, even when seen from the road. Also, Range = u*u*sin(2*theta)/g. So, 60 = 2g*2g*sin(2*theta)/g. So, sin(2*theta) = 15/g. So, theta = 1/2 sin-(15/g).
vinaysajjanapu
18 Points
7 years ago
in the above problem we are getting sin(2theta) value as 3/2 [greater than 1] which is not possible
saraana clinton
12 Points
7 years ago
see ... the ball will now have an horizontal as well as a vertical velocity ..... the horizontal velocity would be the same as the velocity of the truck ... due to inertia .. that is 15m/s and the vertical velocity would be ucos(theta)... hence the answer to the first question would be : as seen by the truck the velocity of the ball will be only the horizontal velocity ... kyuki horizontal direction meh toh truck ki bhi utni hi velocity hai naa... now lets find the vertical velocity .... see its very clear form the que that the range is 60 mt... therefore u*sin2theta =60 .... we know u cos theta =15... use this expand sin2theta and find u sintheta... u will get it as 20 m/s ... now the second part of the question is in reference to the ground ,.... that would be the resultant of usintheta and ucos theta ... that is U .... now we know usintheta and u costheta as well...so definetly its easy to find u .... and we get u as 25 m/s hence....as seen from the ground the velocity of the ball is 25m/s ...... NOTE: * means whole square
saraana clinton
12 Points
7 years ago
i am sorry there was a mistake in my answer... the vertical velocity would be u sin theta not u cos theta ... where theta is the angle of projection... and i am quite a little sure that the remaining part is ri8 !!