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.