Question

A cannonball is fired from the ground and lands on a hill as shown in the figure below: 35° 150 m The initial velocity of the cannonball as it exits the canon is 50 m/s and it is launched at an angle of 35 above the horizontal. The cannonball lands on the hillside 150 m away from the cannon in the horizontal direction (as shown). 2 pts - a) What are the x- and y-components of the initial velocity of the cannonball? 2 pts - b) How long after the cannonball is fired does it land at the point on the hillside (150 m horizontally)? away 3 pts-c)If the cannonball leaves the cannon at y=2.0m above the ground, what is the highest point in the air that the cannonball reaches? (measured above the ground that the cannon sits on. Hint-you'll need to know how long it takes to reach this point) 3 pts -d) What is the angle of elevation of the hill Chint-you'll need to know how high above the starting point the cannonball lands).

Answer 1

(a) Given v_0 = 50 ms^- 1 Horizontal (x - component)of initial velocity: v_0_x = v_0 cos 35^0 = 50 times cos 35^0 = 40.957 ms^- 1 y - component (vertical of velocity: v_0y = v_0 sin 35^0 = 50 times sin 35^0 = 28.678 ms^- 1 (b) The cannon ball travels 150m along x-direction before landing so simply using velocity = distance/time along x-axis will given the time.(since, there is no acceleration along x axis) \r\n implies time = distance/velocity = 150m/v_0x = 150m/40.957ms^- 1 = 3.66s (The velocity in x - direction remains constant through out the motion of the cannon because there is no acceleration in x-direction) (c) If initial launch height is y_0 = 2m, then maximum height reached by the cannon will be: [Using equation of motion: v^2 - u^2 = 2as where v rightarrow final velocity u rightarrow initial velocity a rightarrow acceleration s rightarrow distance travelled] so, in vertical direction: v^2_y - u^2_y = 2gs

Please feel free to ask in case of any doubt and discrepancies.

Please hit a Like.

Thank You.