Acannonball is tired with an initial speed of 40m/s and a launch angle of 30degrees from a cliff that is 25m tall. a. what is the flight time of the cannonball b. how far away does the cannonball land c. using two graphs, plot the horizontal speed of the projectile versus time and the vertical speed of the projectile versus time (from the initial launch of the projectile to the instant i strikes the ground).

Part a)

Part b)

Explanation:

Initial speed of the launch is given as

angle = 30 degree

Now the two components of the velocity

similarly we have

Part a)

Now to find the time of flight we will have

Part b)

Distance at which the ball will land is given as

Part c)

Since x direction there is no acceleration so it will move with uniform speed while in y direction it will accelerate due to gravity

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.