It is possible to shoot an arrow at a speed as high as 100 m / s . (a) If friction is neglected,

It is possible to shoot an arrow at a speed as high as 100 m...

Key Concepts

Conservation of Energy

This concept involves equating the kinetic energy of a moving object to its gravitational potential energy at the peak of its ascent. When friction and air resistance are neglected, the energy at the launch point transforms entirely into potential energy at the maximum height. This allows one to calculate how high an object will rise given its initial speed and the acceleration due to gravity.

Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration, such as gravity. These equations are essential for determining parameters like maximum height and total time of flight when an object is projected straight upwards. They relate initial velocity, acceleration, displacement, and time, and are foundational in solving projectile motion problems.

Time of Flight

Time of flight refers to the total time a projectile remains in the air from launch until it returns to the starting point. In vertical projectile motion, this time can be calculated by determining the time taken to reach the peak using the initial velocity and gravitational acceleration, and then doubling that time, due to the symmetry of the ascent and descent in projectile motion.

Transcript

00:01 Here we have an arrow that can be shot up at a maximum speed of 100 meters per second.

00:04 We are asked to determine how high the arrow can go and for how long it will be in the air.

00:11 We will be ignoring air resistance and the acceleration will be due to gravity alone.

00:17 So we can use the basic kinematic equations such as delta y equals v0 delta t minus one half g delta t squared where i've used a equals negative g.

00:31 Right the acceleration is downward g and we can also use the equation v final squared minus v0 squared equals 2 times a which now is negative g times delta y this is one of my favorite equations here since it doesn't acquire the use of time so we'll actually be using this equation first right since we're asked about delta y and not yet about delta t we're not giving the information about the amount of time so we won't be use that first equation.

01:04 And luckily the second equation has that delta y in it that we can solve for.

01:09 So let's do exactly that and solve for delta y in the second equation.

01:15 We'll have delta y equals v final squared minus v initial squared divided by negative 2g.

01:24 Now the question arises what should be the final velocity? well, we are wondering how how high can this arrow go? well, if you shoot an arrow directly upwards into the sky, it'll go up, and then it'll slow down before it reverses and comes back down.

01:46 So the total height that it can reach will be determined, sorry, it will be determined by a velocity of zero when it reverses direction.

02:00 The velocity becomes zero and then it becomes negative because it's pointing back towards the ground.

02:13 Before that it was positive because it was going upwards.

02:17 So the total delta y will be characterized by a final velocity equal to zero meters per second when the arrow is at a standstill.

02:28 So we have zero minus v0 squared or negative 2g or v0 squared over 2g.

02:36 Is our value of delta y.

02:39 Right, so we have delta y is equal to 100 meters per second squared divided by 2 times 9 .81 meters per second squared.

02:56 Okay, so now we just have to use the calculator.

02:59 So we're going to have one moment.

03:04 We're going to have 100 times 100 divided by 2, divided by 9 .8...

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