In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77 m. What

In 2007, Michael Carter (U.S.) set a world record in the...

Key Concepts

Projectile Launched from an Elevated Position

When a projectile is launched from an elevated position, its trajectory differs from that of an object launched from ground level. The initial height affects the time of flight and the horizontal distance traveled. Understanding this concept is important when applying the kinematic equations, as the vertical displacement will involve an initial height term leading to a quadratic equation that must be solved to determine the overall motion parameters.

Kinematic Equations

Kinematic equations are the set of formulas used to describe the motion of objects under constant acceleration, such as gravity. These equations relate displacement, initial velocity, time, and acceleration, allowing one to compute unknown variables like the initial speed required given a specific range, launch height, and angle of projection.

Projectile Motion

Projectile motion describes the two-dimensional motion of an object subject only to the force of gravity. When an object is projected into the air, its motion can be separated into independent horizontal and vertical components. This concept is fundamental for understanding trajectories, range, time of flight, and maximum height reached by the object.

Decomposition of Initial Velocity

The decomposition of initial velocity involves breaking the velocity vector into its horizontal and vertical components using trigonometric functions. When an object is launched at an angle, the horizontal component (using cosine) and the vertical component (using sine) are used to analyze the motion independently. This approach is essential for solving problems involving trajectories and determining parameters such as range and flight duration.

Transcript

00:03 For this example, we want to determine the initial velocity of a ball thrown from a height of a man standing at 2 .1 meters tall.

00:14 Given that we throw this ball at an initial angle of 38 degrees, and we know that it travels a total distance of 24 .77 meters.

00:26 So in order to do this, first we're going to analyze this projectile problem in the horizontal and vertical parts.

00:36 Starting with the horizontal piece.

00:40 So in the x direction, we know that the horizontal component of velocity will never change during this throw because only the vertical component of velocity will change on the surface of the earth because of the force of gravity.

00:55 And we know that the distance traveled by the ball x should just obey the relation x equals x not plus the x component of velocity times time.

01:10 And of course x not, we're just defining as zero in our coordinate system because we're anchoring the origin of our coordinate system right here.

01:18 So this is plus x.

01:20 This is plus y.

01:26 So really we just have the relationship for constant velocity here, which is that velocity is distance over time.

01:38 So notice if we plug in our expression for the x component of velocity is v0 times the cosine of the initial angle we throw it at we end up with x equals v not cosine of theta times t so we have an equation here where the only thing we're missing in order to solve this for v not is just time so we have two unknowns here so if we could go deal with one of our unknowns then we could go ahead and solve for v0 so we're going to investigate the vertical direction of motion now and see if we can get another equation with time as an unknown.

02:35 So given that we're using initial and final y positions and we also have like an intermediate expression for our vertical velocity, the equation i'm going to go to, or the kinematic equation i'm going to go to in the y direction is going to be y minus y not equals v0 y times t plus one half a t squared where a is just going to be minus the acceleration due to gravity on earth's surface.

03:08 We'll have minus one -half gt squared here.

03:13 So this equation definitely has time as an unknown, so that's promising.

03:20 We know what the left -hand side is because we have y in the final height and the initial height.

03:26 But we'll have to break down this initial y component of velocity a little bit here and see what we can do.

03:34 So again, we're just going to use our other trigonometric expression for our components of velocity, which is going to be that the initial y component of velocity is just the initial velocity times the sign of the angle we threw it out.

03:53 So we'll have v0 times the sine of theta times t minus one half gt squared, and then the left side is the same.

04:06 Okay, so let's look at these two equations.

04:09 So we'll call this equation two, and i'll call this first one up here.

04:12 Equation one.

04:15 And in fact, let's just rewrite equation one down here so that we can have them side by side and then figure out a strategy.

04:27 So equation one was just that x is equal to b .0 cosine of theta times time.

04:38 Pretty sure.

04:38 Yeah.

04:40 Okay.

04:42 So both of these equations have v not as an unknown, which is what we want to solve for.

04:47 And they both have time as an unknown.

04:51 So what we're going to do here is just a classic substitution with these two equations.

05:04 So we just have to decide how we want to do this substitution.

05:07 And the way i think we should do it is since we're eliminating time, we'll take equation one here, solve for time, and then plug that into equation two...

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