A home run is hit in such a way that the baseball just...

A home run is hit in such a way that the baseball just clears a wall $21.0 \mathrm{~m}$ high, located $180 \mathrm{~m}$ from home plate. The ball is hit at an angle of $35.0^{\circ}$ to the horizontal, and air resistance is negligible. Find (a) the initial speed of the ball, (b) the time it takes the ball to reach the wall, and (c) the velocity components and the speed of the ball when it reaches the wall. (Assume the ball is hit at a height of $1.00 \mathrm{~m}$ above the ground.)

Key Concepts

Projectile Motion

Projectile motion concerns the motion of an object that is launched into the air and then subject only to the acceleration due to gravity. In answering the problem, this concept underpins how both the horizontal and vertical motions of the baseball are analyzed separately, revealing that horizontal motion occurs at constant velocity while vertical motion is uniformly accelerated.

Decomposition of Velocity

When an object is launched at an angle, its initial velocity must be broken down into horizontal and vertical components using trigonometric functions. This decomposition is essential to apply the appropriate kinematic equations to each direction independently, allowing for the calculation of distances, times, and speeds in both the horizontal and vertical dimensions.

Kinematic Equations for Constant Acceleration

Kinematic equations describe the motion of objects under constant acceleration, such as gravity. These equations are used to relate displacement, initial velocity, time, and acceleration, and are crucial for solving for unknown quantities like the initial speed and time taken for the projectile to reach a specific point.

Determining Time of Flight to a Specific Point

The time of flight to the wall is determined by analyzing the horizontal motion, where the displacement and constant horizontal velocity provide a direct method to calculate the time it takes for the baseball to reach the wall. This time is then used in the vertical motion analysis to ensure that the ball just clears the wall.

Vertical Displacement and Final Velocity

The vertical displacement equation, which includes the effects of gravitational acceleration, helps track the change in vertical position from the point of launch to the point where the ball reaches the wall. It also aids in calculating the vertical velocity at that point by considering the influence of gravity over the elapsed time.

Transcript

0:00 Hello.

00:02 In this problem, we have a baseball ball that has been shot from a place 180 meters away from a wall that is 21 meters in height.

00:12 And the ball was shot from one meters above the ground at an angle of 35 degrees above the horizon.

00:22 We're asking this problem to calculate the initial velocity such that the ball just clears the wall when it reaches to this position.

00:32 So here again let me just put our numbers on the screen.

00:38 The angle is 35 degrees and the unknown is vi.

00:43 So the first point as we said is to calculate the initial velocity.

00:48 I will start by writing down the expression for the x coordinate of the trajectory at this point.

00:54 So vi cosign 35 as usual, multiply by t.

01:01 And at this point the ball was 180 meters away from the point.

01:04 Its starting point.

01:06 From here we can get a simplified expression for time, which is after simplifying and divided by cosine 35, this gives us to 12, 219 .7 divided by the initial velocity that we are interested to calculate.

01:23 Now let's start by writing the same expression by, but this time for the height or the y coordinate at this time.

01:31 So the equation for the height is v .i, sign the angle, which is 35 in that case, multiply by time, minus half the gravitational acceleration, which is 4 .9, t squared.

01:45 Now let's substitute using this expression for time in this equation here.

01:52 So this will give us v .i.

01:56 Sign 35 multiplied by t, putting in t as 219 .7 over vi...

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