The velocity-time graph of a block projected up a smooth inclined plane is a straight line. The block starts with an initial velocity of 9.8 m/s at t=0 and momentarily comes to rest at t=2 s . Taking g = 9.8 m/s ^2 , the angle of inclination of the plane is:

Step-by-step solution

From the given velocity-time data, the magnitude of deceleration is: a = v t = 9.8 - 0 2 - 0 = 4.9 m/s ^2 For a block moving up a smooth inclined plane of angle , the deceleration is provided entirely by the component of gravity along the incline: a = g Equating the two expressions for deceleration: g = 4.9 9.8 = 4.9 = 4.9 9.8 = 1 2 Therefore, the angle of inclination is: = 30^ Answer: 30^