A cannonball is shot (from ground level) with an initial horizontal velocity of 41 m/s and an initial vertical velocity of 22 m/s.
1) What is the speed of the cannonball 1.7 seconds after it was shot?
2) How high above the ground is the cannonball 1.7 seconds after it is shot?
1) the horizontal velocity stays constant
... the vertical velocity is decreased by gravitational acceleration ... 9.8 m/s^2
... 22 - (1.7 * 9.8)
... add the two velocity vectors to find the resultant
2) you have the vertical velocity at 1.7 s
... average it with the initial (22 m/s)
... multiply the average by 1.7 s to find the height
Vo = Xo + Yo = 41 + 22i m/s.
1. V = Y = Yo + g+t = 22 - 9.8*1.7 = 5.34 m/s.
Vo = Xo + Yi = 41 + 5.34i = (Convert to polar form).
2. Y^2 = Yo^2 + 2g*h.
5.34^2 = 22^2 - 19.6h, h = ?.
To answer these questions, we can use the equations of motion to analyze the horizontal and vertical components of the cannonball's motion separately.
Given:
Initial horizontal velocity (Vx) = 41 m/s
Initial vertical velocity (Vy) = 22 m/s
Time (t) = 1.7 seconds
1) To find the speed of the cannonball 1.7 seconds after it was shot, we need to calculate the resultant velocity (V) using the horizontal (Vx) and vertical (Vy) velocities.
The speed of the cannonball can be found using the Pythagorean theorem:
V = √(Vx² + Vy²)
Substituting the given values:
V = √(41² + 22²)
V = √(1681 + 484)
V = √(2165)
V ≈ 46.54 m/s
Therefore, the speed of the cannonball 1.7 seconds after it was shot is approximately 46.54 m/s.
2) To determine the height above the ground after 1.7 seconds, we can use the vertical motion equation:
h = Vy * t + (1/2) * g * t²
where:
h is the vertical displacement or height
t is the time (1.7 seconds in this case)
Vy is the initial vertical velocity (22 m/s)
g is the acceleration due to gravity (9.8 m/s²)
Substituting the given values:
h = 22 * 1.7 + (1/2) * 9.8 * 1.7²
h = 37.4 + 14.483
h ≈ 51.88 meters
Therefore, the cannonball is approximately 51.88 meters above the ground 1.7 seconds after it is shot.