An arrow is shot from a height of 1.5 m toward a cliff of...

An arrow is shot from a height of 1.5 m toward a cliff of height H . It is shot with a velocity of 30 m/s at an angle of $60^{\circ}$ above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow's impact speed just before hitting the cliff?

An arrow is shot from a height of 1.5 m toward a cliff of height H . It is shot with a velocity of 30 m/s at an angle of $60^{\circ}$ above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c)
What is the arrow's impact speed just before hitting the cliff?

Key Concepts

Projectile Motion

Projectile motion describes the two-dimensional trajectory of an object that is launched into the air and influenced only by the acceleration due to gravity. This motion is typically analyzed by breaking it into horizontal and vertical components that evolve independently, with the horizontal motion occurring at a constant speed and the vertical motion undergoing constant acceleration.

Decomposition of Velocity Vectors

Decomposing the initial velocity into horizontal and vertical components is crucial in projectile motion analysis. The horizontal component is determined by multiplying the initial speed by the cosine of the launch angle, while the vertical component is found by multiplying the initial speed by the sine of the launch angle. This separation allows for independent analysis of motion in each direction.

Kinematic Equations

The kinematic equations govern the motion of objects experiencing constant acceleration. They are used to relate displacement, velocity, time, and acceleration in both the horizontal and vertical directions. These equations provide the mathematical framework to predict the position and velocity of a projectile at any given time.

Maximum Height Calculation

The maximum height reached by a projectile occurs when the vertical component of its velocity becomes zero. By applying the kinematic equations to the vertical motion and setting the final vertical velocity to zero, one can calculate the maximum vertical displacement from the launch point, which is a key aspect of projectile motion analysis.

Impact Speed Determination

Calculating impact speed in a projectile motion scenario involves determining the velocity components just before impact and then combining them to find the resultant speed. The horizontal component remains constant, while the vertical component is found using the kinematic equations to account for the acceleration due to gravity. The overall impact speed is the magnitude of the vector sum of these two components.

Transcript

00:01 So here for part a, we can find the height of the cliff.

00:02 We can say the height is equaling y initial plus v.

00:06 Y initial t plus one half gt squared.

00:10 So here we have that the height of the cliff is equaling 1 .5 meters plus 30 meters per second times sign of 60 degrees multiplied by t.

00:24 So four seconds plus rather we could say plus plus one.

00:30 1 1⁄2 times negative 9 .80 meters per second squared, multiplied by 4 seconds quantity squared, and the height of the cliff is found to be 27 .0 meters.

00:43 This would be our final answer for part a.

00:46 For part b now, they want us to find the maximum height, and we know that here, the maximum height, for part b at the maximum height, the y component of the velocity, or the final y component of the velocities is equaling zero meters per second.

01:07 So we can say vy final squared equals vy initial squared plus 2g times y max.

01:16 Given that here, essentially we can say that the y the height reached y initial is zero meters this is considered essentially right at the cliff so we can say that the maximum height would be y max equaling negative v y initial square divided by two g this is equaling negative 30 meters per second times sign of 60 degrees this term would be square this term would be squared divided by 2 times negative 9 .8 meters per second squared and so this is equaling 34 .44 meters take note of this number and then for part b when it's asking for the maximum height reached we can say max we can say rather nor differentiate h max this would be equal to y max plus the height of the cliff itself so y max plus 1...

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