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Olympic skier jumpers use their own bodies as projectiles.
Projectile motions are a projectile under gravity's motion
Projectiles are anything moving due to gravity with 2D curved trajectories.
E.g. 2 person each have identical balls that simultaneously, one their ball (Ball 1) throws horizontally, while the other drops their ball (Ball 2) to the ground: Both balls reach the ground at the exact ame time.
Recall the river crossing problems: the boat experiences 2 velocities perpendicular to each other. Since the velocity across the river and down the river have no effect on each other, these 2 are independent.
Time of flight is how long it takes for something to complete its motion.
Earlier, both balls experienced vertical acceleration due to gravity.
Though Ball 1 is projected horizontally, its horizontal doesn't affect its vertical motion, as projectile's horizontal is perpendicular to its vertical velocity.
These 2 velocities are each independent.
Figure 2 shows that horizontal velocity is constant, while vertical velocity increases from 0 with uniform acceleration.
Ball 1 undergo same vertical motion as Ball 2 falling straight. Result: both reach the ground at same time.
But the river and the projectile motion (balls) problem has a difference: River problem has both velocities constant.
The projectile motion problem has its horizontal velocity constant and its vertical velocity changes as of acceleration due to gravity.
The balls undergo uniform velocity horizontally and uniform acceleration vertically.
These 2 motions are independent, but both have time.
Time taken for horizontal motion is the same as the one taken for vertical motion since projectile rest when hitting ground. The range (Δdx) is a horizontal distance travelled by a projectile.
Describing projectile motion
Describing projectile motion
Projectile motion problems = 2D dimensional vector problems. To describe motion in this type of problem via vector, use the convention that velocity vectors pointing upward or to the right have positive values and velocity vectors pointing downward or to the left have negative values.
A method used here to solve projectile motion problems is to work with motion in only one direction (horizontal or vertical) at a time, with the info provided about motion in a direction to solve for time value.
Since we work with only 1 motion at any given time, vector arrows (→) aren't used here.
But recall that projectile motion problem = vector problems.
→ →
When something is launched, in horizontally, its initial velocity is vix and vertically is viy = 0
Problem 1
Problem 1
A bag is thrown out of a window 10.0 above ground with an initial horizontal velocity of 3.0 m/s.
a) How long will it take the bag to reach ground/what is its time of flight?
b) How far will the bag travel horizontally/what is its range?
Solution:
a) To find the time of flight, consider its vertical motion.
given: Δdy = -10 n; ay = -9.8/s²; viy = 0 m/s
required: Δd
analysis: We can use one of 5 motion equations of Section 1.5 to solve for time taken to bag to reach the ground:
See how the bag undergoes motion in a parabola shape and that the statement says that the vertical displacement is negative, showing that the bag is falling.
Similarly acceleration due to gravity is given negative.
The initial velocity in vertical direction is zero as the bag is not thrown up or downard.
b) To find the bag's horizonatl distance/range, consider its horizontal motion. We use the fact that both motions, vertical and horizontal, the the same amount of time.
given: Δt = 1.43 s; vix = 30 m/s; ax = 0 m/s²
required: Δdx = vix Δt
analysis: Δdx = vix Δt
= (3.9 m.s)(1.43 s)
Δdx = 4.3 m
The bag travels 4.3 horizontally.
Problem 2
Problem 2
In golf, Joe tries to improve his shot's range, by driving his ball from the top of a steep clif, 30.0 m above the ground where the ball will land.
If the ball has an initial velocity of 25 m/s and is launched at an angle of 50° above the horizontal determine its time of flight, range, and final velocity before hitting ground.
Figure 8 shows the golf ball's motion.
Notes: Here we'll combine the horizontal and vertical given statements.
Figure 8
Solution:
See that this is a quadratic equation for time. Earlier, when solving this equation for time, a variable in here equals to zero, allowing to solve for Δt without solving a quadratic relation. Here we must expand the equation and use the quadratic formula.
Notes: Ignore the units in the end to simplify calculations.
Statement: The ball's range = 82 m. To determine its final velocity just before touching ground, consider Figure 9 (right), showing its final velocity and its horizontal and vertical components.
Since the ball travels at constant horizontal velocity, we know that
the final horizontal velocity (vfs) = initial horizontal velocity (vix).
But in the vertical direction, the initial and final velocities aren't the same as the ball accelerates vertically.
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