A cannonball is fired with an angle of 40 degree above the horizontal.If the velocity when it hits the ground is 80 m/s
FIND
a.Viy b.time of travel c.max height d.range
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To find the required quantities, we'll use the equations of projectile motion.
Given:
Initial angle of the cannonball (θ) = 40 degrees
Final velocity when it hits the ground (Vf) = 80 m/s
a. Finding the initial vertical velocity (Viy):
The initial velocity (Vi) can be split into its horizontal and vertical components:
Vi = Vix + Viy
Since the cannonball is fired horizontally, the initial horizontal velocity (Vix) is given by:
Vix = Vi * cos(θ)
To find the initial vertical velocity (Viy), we'll use the equation:
Viy = Vi * sin(θ)
Plugging in the values:
Vix = Vi * cos(40°)
Viy = Vi * sin(40°)
b. Finding the time of travel (t):
To find the time of travel, we can first find the time it takes for the cannonball to reach its maximum height (t_peak), and then double it to get the total time of travel (t_total).
The time it takes for the cannonball to reach its maximum height is given by:
t_peak = Viy / g
Where g is the acceleration due to gravity (9.8 m/s²).
The total time of travel is given by:
t_total = 2 * t_peak
c. Finding the maximum height (H):
The maximum height can be calculated using the equation:
H = (Viy²) / (2 * g)
d. Finding the range (R):
The range is the horizontal distance traveled by the cannonball. It can be calculated using the equation:
R = Vix * t_total
Now let's calculate the values using the given information:
a. Viy:
Viy = Vi * sin(40°)
b. t_total:
t_peak = Viy / g
t_total = 2 * t_peak
c. H:
H = (Viy²) / (2 * g)
d. R:
R = Vix * t_total
Plug in the values to find the answers.