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Olin [163]
3 years ago
12

A cannonball 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 t

he 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).

Physics
1 answer:
ella [17]3 years ago
7 0

Answer:

Part a)

t = 5.1 s

Part b)

x = 176.7 m

Explanation:

Initial speed of the launch is given as

v_i = 40 m/s

angle = 30 degree

Now the two components of the velocity

v_x = v_i cos30

v_x = 34.6 m/s

similarly we have

v_y = v_i sin30

v_y = 20 m/s

Part a)

Now to find the time of flight we will have

\Delta y = v_y t + \frac{1}{2}at^2

-25 = 20 t -\frac{1}{2}(9.81)t^2

t = 5.1 s

Part b)

Distance at which the ball will land is given as

x = v_x t

x = 34.6 (5.1)

x = 176.7 m

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

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