| BETA : Translation : Projectile Motion 01 : Vertical Projectile Motion Returning To Same Elevation : BETA Physics Homework Exercise
Description : Vertical projectile motion returning to same elevation.
Discussion : A projectile fired vertically will reach a maximum height before accelerating back towards the ground at the starting point. The trajectory is due to gravitational acceleration. Friction is ignored. See Figure Vertical Projectile Motion Due To Gravity
The homework exercise has 3 sets of questions with 2 questions per set.
h, t
t, vi
h, vi
The basic equations are :
h = vi2 / (2 g)
t = √(8 h / g)
rearranging
t = √(8 h / g)
h = g t2 / 8
vi = t g / 2
vi = √(2 g h)
where
vi = initial velocity (vertical)
t = time
h = maximum height
g = gravity constant = 9.81 m / s2
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| BETA : Translation : Projectile Motion 02 : Vertical Projectile Motion Returning To Different Elevation : BETA Physics Homework Exercise
Description : Vertical projectile motion returning to different elevation
Discussion : A projectile fired vertically will reach a maximum height before accelerating back towards the ground at the starting point. The final elevation is different to the initial elevation. The trajectory is due to gravitational acceleration. Friction is ignored. See Figure Vertical Projectile Motion Due To Gravity
The homework exercise has 3 sets of questions with 2 questions per set.
h, t
t, vi
h, z
The basic equations are :
h = vi2 / (2 g)
t = √(2 h / g) + √(2 (h - z) / g)
rearranging
vi = √(2 g h)
z = h - g / 2 (t - √(2 h / g))2
where
vi = initial velocity (vertical)
t = time
h = maximum height
z = final elevation relative to datum
g = gravity constant = 9.81 m / s2
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| BETA : Translation : Projectile Motion 03 : Angled Projectile Motion Returning To Same Elevation : BETA Physics Homework Exercise
Description : Angled projectile motion returning to the same elevation.
Discussion : A projectile fired upwards at an angle will follow a curved trajectory, reaching a maximum height before accelerating towards the ground some distance from the starting point. The trajectory is due to gravitational acceleration. The horizontal component of velocity is constant. Friction is ignored. The ground elevation at the final position is the same as the initial elevation (datum). See Figure Angled Projectile Motion Due To Gravity
The homework exercise has 3 sets of questions with 5 questions per set.
vv, vh, t, d, h
vv, vh, h, vi, theta
t, vh, vv, vi, theta
The basic equations are :
vv = vi sind(theta)
vh = vi cosd(theta)
t = 2 vv / g
d = vh t
h = vv2 / (2 g) = g t2 / 8
d = vh t
rearranging
vv = t g / 2
vh = d / t
t = d / vh
t = √(8 h / g)
vi = √(vv2 + vh2)
theta = atan2d(vv, vh)
where
vi = initial velocity
vv = vertical velocity component
vh = horizontal velocity component
t = time
d = horizontal distance
h = maximum height
theta = angle of elevation (θ)
g = gravity constant = 9.81 m / s2
The angle theta (θ) is in degrees. you are recommended to use the trigonometry degree functions sind(0, cosd(0 and tan2d().
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| BETA : Translation : Projectile Motion 04 : Angled Projectile Motion Returning To Different Elevation : BETA Physics Homework Exercise
Description : Angled projectile motion returning to different elevation
Discussion : A projectile fired upwards at an angle will follow a curved trajectory, reaching a maximum height before accelerating towards the ground some distance from the starting point. The trajectory is due to gravitational acceleration. The horizontal component of velocity is constant. Friction is ignored. The ground elevation at the final position is different to the initial elevation (datum). See Figure Angled Projectile Motion Due To Gravity
The homework exercise has 3 sets of questions with 4 questions per set.
vv, vh, h, t, d
h, t, z, vi, theta
vv, t, vh, d, theta
The basic equations are :
vv = vi sind(theta)
vh = vi cosd(theta)
h = vv2 / (2 g)
t = √(2 h / g) + √(2 (h - z) / g)
d = vh t
rearranging
t = d / vh
z = h - g / 2 (t - √(2 h / g))2
vi = &radic(vv2 + vh2)
vh = &radic(vi2 - vv2)
theta = atan2d(vv, vh)
where
vi = initial velocity
vv = vertical velocity component
vh = horizontal velocity component
t = time
d = horizontal distance
h = maximum height
theta = angle of elevation (θ degrees)
z = final elevation relative to the datum
g = gravity constant = 9.81 m / s2
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| BETA : Translation : Projectile Motion 05 : Horizontal Projectile Motion To Lower Elevation : BETA Physics Homework Exercise
Description : Horizontal projectile motion to lower elevation.
Discussion : A projectile fired horizontally will follow a curved trajectory accelerating towards the ground some distance from the starting point. The trajectory is due to gravitational acceleration. The horizontal component of velocity is constant. Friction is ignored. See Figure Horizontal Projectile Motion Due To Gravity
The homework exercise has 3 sets of questions with 3 questions per set.
t, d, vv
vi, h, vv
t, vi, h
The basic equations are :
t = √(2 h / g)
d = vi t
vv = g t
rearranging
vi = d / t
h = 1/2 g t2
vv = g t
t = vv / g
where
vi = initial velocity
vv = final vertical velocity component
t = time
d = horizontal distance
h = initial height
g = gravity constant = 9.81 m / s2
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