Final answer:
The initial horizontal velocity (u) of the cannonball is 7 m/s, determined by the time it takes to fall 122.5m vertically due to gravity. The initial vertical velocity (V) is 0 m/s because the cannon is shot horizontally.
Step-by-step explanation:
The student's question relates to projectile motion where a cannonball is shot horizontally from a cliff, and we need to determine the initial horizontal and vertical velocities. Since the cannonball lands 35m away from the base of the cliff and the cliff is 122.5m tall, the problem can be solved using the kinematic equations for projectile motion.
To calculate initial horizontal velocity (u), consider that this velocity will remain constant throughout the flight of the projectile, as there is no horizontal acceleration. The time (t) it takes for the projectile to hit the ground only depends on the vertical motion. The vertical distance (122.5m) and acceleration due to gravity (9.8 m/s2) can be used to find the time.
Using the equation s = ut + ½at2, where s is the vertical displacement, u is initial vertical velocity (which is 0 since it is shot horizontally), a is the acceleration due to gravity, and t is the time, we can solve for t as follows:
0 = 0×t + ½×9.8×t2
The time of flight can be calculated for the vertical movement by solving:
122.5 = ½×9.8×t2 → t = sqrt(122.5×2/9.8) → t = √(25) → t = 5s.
Therefore, to find the initial horizontal velocity, use the equation s = ut, where distance s is 35m and time t is 5s:
35 = u×5 → u = 35/5 → u = 7 m/s.
The initial vertical velocity (V) is 0 m/s since the cannonball is shot horizontally. Thus, the correct answer is (a) u = 7 m/s, V = 0 m/s.