Two seconds after being projected from ground level, a...

Two seconds after being projected from ground level, a projectile is displaced $40 \mathrm{~m}$ horizontally and $58 \mathrm{~m}$ vertically above its launch point. What are the (a) horizontal and (b) vertical components of the initial velocity of the projectile? (c) At the instant the projectile achieves its maximum height above ground level, how far is it displaced horizontally from the launch point?

Key Concepts

Projectile Motion

Projectile motion refers to the two-dimensional movement of an object that is launched into the air and moves under the influence of gravity alone. The motion is typically decomposed into independent horizontal and vertical components, with gravity affecting only the vertical motion and the horizontal motion continuing at a constant velocity (in the absence of air resistance).

Decomposition of Vectors

Decomposition of vectors involves breaking down the initial velocity into its horizontal and vertical components. This decomposition is essential for analyzing projectile motion, as it allows separate treatment of horizontal motion (with constant velocity) and vertical motion (with constant acceleration due to gravity).

Kinematic Equations

Kinematic equations are used to describe the motion of objects under constant acceleration. For projectile motion, the equations allow us to compute parameters such as displacement, velocity, and time in both the horizontal (constant velocity) and vertical (uniformly accelerated motion) directions. These equations are fundamental tools for solving problems related to projectile trajectories.

Acceleration Due to Gravity

Acceleration due to gravity is the constant acceleration experienced by an object in free fall near the Earth's surface. In projectile motion, this acceleration acts only in the vertical direction, affecting the object's upward and downward motion. It plays a crucial role in determining the time it takes for the projectile to reach its highest point and to return to the ground.

Maximum Height and Horizontal Displacement

At maximum height, the vertical component of the projectile's velocity becomes zero. The time to reach this point can be determined using kinematic equations, and it can then be used to calculate how far the projectile has traveled horizontally. This concept is critical in understanding the overall trajectory and the range of the projectile.

Transcript

00:01 This is chapter 4, problem number 50.

00:03 We have a projectile in 2 equals 2 seconds.

00:08 T equals 2 seconds.

00:10 The range, the horizontal displacement, it undergoes is 40 meters.

00:15 And the height that it reaches is 53 meters.

00:19 So in part a, we're asked the x component of the initial velocity under the circumstances.

00:27 Acids.

00:27 So since there is no acceleration in the x direction, the range travel is going to be equal to v -n -x times time that it takes to reach that horizontal displacement.

00:38 So it's pretty straightforward here.

00:40 Then if we divide both sides by time, we're getting the x component of the initial velocity as are over time, r being 40 meters and time being two seconds, we find the x component of the velocity to be 20 meters per second.

00:55 In part b, we're asked, velocity's y component this time so in order to do that we're going to take advantage of what's given to us so delta y is given as v0y times t minus one half gt squared delta y being our h here that is given to us as 53 meters so then i'm going to solve this equation for v0 y so we not y times time equals h plus one half gt squared if i divide both sides by t right one of our t, then i'm going to get v .0 .y.

01:30 So v .0 .y equals h being 53 meters plus 1 .5, 9 .8 meters per second square times, time that is given to us as 2 seconds squared.

01:43 So big brackets, 1 over 2 seconds...

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