Kinematics. cannonball was dropped- how far from the wall does it hit ground?

[aurorabrv]

Homework Statement

King Arthur's knights fire a cannon from the top of the castle wall. The cannonball is fired at a speed of 45m/s and an angle of 32degrees . A cannonball that was accidentally dropped hits the moat below in 1.4s .

a) How far from the castle wall does the cannonball hit the ground?

b) What is the ball's maximum height above the ground?

Homework Equations

Unsure what to use.

The Attempt at a Solution

xi = ti = 0
vi = 45 m/s
vix = 45cos32 = 38.16
viy = 45sin32 = 23.85
ax = 0
ay = -9.8 m/s
yf = 0
yi = 1/2 (-9.8m/s^2)(1.4s)^2 = 9.6 m

horizontal v 45cos32 = 38.16
initial v 45sin32 = 23.85

time of fall 2.43 seconds

57.86/9.8 = root of 5.9
= 2.43 + 2.43
= 4.86
= 185.4576m


max height
1/2(9.8) x 2.43^2 = 28.93 m

 

[Defennder]

Wow, I can't believe this problem is almost the same as this one:
https://www.physicsforums.com/showthread.php?t=253840

 

[baba89]

For this problem, we can use the equations of kinematics to calculate the distance and maximum height of the cannonball. We know that the initial velocity of the cannonball is 45m/s at an angle of 32 degrees. Using this information, we can break down the velocity into its horizontal and vertical components, as shown in the attempt at a solution.

To calculate the distance from the castle wall, we can use the equation x = x0 + v0t + 1/2at^2, where x0 is the initial position, v0 is the initial velocity, a is the acceleration, and t is the time. Since the cannonball is dropped, its initial velocity is 0, and the acceleration is -9.8m/s^2 due to gravity. Plugging in the values, we get:

x = 0 + 0 + 1/2(-9.8m/s^2)(1.4s)^2 = 9.6m

Therefore, the cannonball hits the ground 9.6m away from the castle wall.

To find the maximum height, we can use the equation y = y0 + v0t + 1/2at^2, where y0 is the initial height, v0 is the initial velocity, a is the acceleration, and t is the time. Since the cannonball reaches its maximum height at the peak of its trajectory, we know that its final height is 0. Plugging in the values, we get:

0 = y0 + 23.85m/s(2.43s) + 1/2(-9.8m/s^2)(2.43s)^2

Solving for y0, we get y0 = 28.93m.

Therefore, the maximum height of the cannonball is 28.93m above the ground.