A projectile is fired from ground level with an initial velocity of 50m/s and an initial angle of 30degree. Assuming g= 9.8 m/s*2, find:
(a) The projectile total time of flight.
(b) The maximum height attained
(c) The total horizontal traveled
(d) The final horizontal and vertical velocities just before it hits the ground
t= 2vₒ•sinα/g,
h= vₒ²•sin²α/2g,
L=vₒ²•sin2α/g,
v(fin) =v(init)
v(x)=v(x0) =v(0)•cosα
v(y)=-v(y0)=-v(0)•sinα
To find the answers to these questions, we can use the equations of projectile motion. Let's break it down step by step:
(a) The projectile total time of flight:
To find the total time of flight, we need to calculate the time it takes for the projectile to reach its highest point, and then double that time.
We can start by calculating the time it takes for the projectile to reach its highest point using the vertical motion equation:
h = (v0y^2)/(2g)
Where:
h = maximum height attained
v0y = initial vertical velocity
g = acceleration due to gravity
Given that the initial vertical velocity (v0y) is 50m/s and the acceleration due to gravity (g) is 9.8m/s^2, we can substitute these values into the equation:
h = (50^2)/(2 * 9.8)
Calculate h:
h = 127.55m
Now, to find the total time of flight, we double the time it took to reach the maximum height:
t = 2 * (v0y/g)
t = 2 * (50/9.8)
Calculate t:
t = 10.20s
So, the total time of flight is approximately 10.20 seconds.
(b) The maximum height attained:
We have already calculated the maximum height (h) in the previous step:
h = 127.55m
So, the maximum height attained is approximately 127.55 meters.
(c) The total horizontal distance traveled:
To find the total horizontal distance traveled, we can use the equation:
R = v0x * t
Where:
R = total horizontal distance traveled
v0x = initial horizontal velocity
t = total time of flight
Given that the initial horizontal velocity (v0x) is calculated by v0x = v0 * cos(θ), where v0 is the initial velocity (50m/s) and θ is the initial angle (30 degrees).
Substitute the values into the equation:
v0x = 50 * cos(30)
Calculate v0x:
v0x = 43.30 m/s (approximately)
Now, substitute the values into the equation for total horizontal distance:
R = v0x * t
R = 43.30 * 10.20
Calculate R:
R = 442.26 m
So, the total horizontal distance traveled is approximately 442.26 meters.
(d) The final horizontal and vertical velocities just before it hits the ground:
The final horizontal velocity remains constant throughout the motion, while the final vertical velocity can be found using the equation:
vf = v0y - g * t
Where:
vf = final vertical velocity
v0y = initial vertical velocity
g = acceleration due to gravity
t = total time of flight
Given that v0y is 50 m/s, g is 9.8 m/s^2, and t is 10.2 s (from the previous calculations), we can substitute these values into the equation:
vf = 50 - (9.8 * 10.2)
Calculate vf:
vf = -49.96 m/s (approximately)
Note: The negative sign indicates that the final vertical velocity is in the opposite direction of the initial vertical velocity, which means the projectile is moving downward.
So, the final horizontal velocity remains approximately 43.30 m/s, and the final vertical velocity is approximately -49.96 m/s.