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Projectile motion practice, Summaries of Advanced Physics

Nueva Vizcaya State University (NVSU)Advanced Physics

sand trip with an initial velocity of 45 feet per second. What is the maximum height that the golf ball will reach? Page 2. Solution details for ...

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2021/2022

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Projectile motion practice
Let’s solve the example of a quadratic equation involving maximums and minimums for projectile motion
1. A ball is thrown directly upward from an initial height of 200 feet with an initial velocity of 96 feet per
second. After how many seconds will the ball reach its maximum height? And, what is the maximum
height? (detailed solution available for this problem)
2.
Some fireworks are fired vertically into the air from the ground at an initial velocity of 80 feet per second. Find
the highest point reached by the projectile.
3.
A ball is thrown vertically upward with an initial velocity of 48 feet per second. If the ball started from
a height of 8 feet off the ground, determine the time it will take for the ball to hit the ground.
4. Multiple choice!
A pistol is accidentally discharged vertically upward at a height of 3 feet above the ground. If the
bullet has an initial muzzle velocity of 200 feet per second, what maximum height will it reach
before it starts to fall to the ground?
628 feet
1,878 feet
199.33 feet
20.87 feet
5.
An over zealous golfer hits a flop shot (ball will go straight up) with a sand wedge to get out of the corner of a
sand trip with an initial velocity of 45 feet per second. What is the maximum height that the golf ball will
reach?
pf3
pf4

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Projectile motion practice

Let’s solve the example of a quadratic equation involving maximums and minimums for projectile motion

  1. A ball is thrown directly upward from an initial height of 200 feet with an initial velocity of 96 feet per second. After how many seconds will the ball reach its maximum height? And, what is the maximum height? (detailed solution available for this problem)
  2. Some fireworks are fired vertically into the air from the ground at an initial velocity of 80 feet per second. Find the highest point reached by the projectile.
  3. A ball is thrown vertically upward with an initial velocity of 48 feet per second. If the ball started from a height of 8 feet off the ground, determine the time it will take for the ball to hit the ground.
  4. Multiple choice! A pistol is accidentally discharged vertically upward at a height of 3 feet above the ground. If the bullet has an initial muzzle velocity of 200 feet per second, what maximum height will it reach before it starts to fall to the ground? 628 feet 1,878 feet 199. 33 feet 20. 87 feet
  5. An over zealous golfer hits a flop shot (ball will go straight up) with a sand wedge to get out of the corner of a sand trip with an initial velocity of 45 feet per second. What is the maximum height that the golf ball will reach?

Solution details for problem # To analyze our problems, we will be using a formula for a freely falling body in which we can ignore any effects of air resistance.  s(t) represents the projectile's instantaneous height at any time t  vo represents initial velocity  so represents the initial height from which the projectile is released  t represents time in seconds after the projectile is released In this formula, -16 is a constant is based on the gravitational force of the earth and represents ½ g = ½(-32 ft/sec^2 ) = -16 ft/sec^2. Since g , or the acceleration due to gravity, is being measured in ft/sec^2 , we must also measure s(t) , vo , and so in terms of feet and seconds. Let's begin by substituting known values for variables in the formula in problem #1: Since the formula represents a parabola, we must find the vertex of the parabola to find the time it takes for the ball to reach its maximum height as well as the maximum height (called the apex). Using the vertex formula: seconds Substituting into the projectile motion formula we have: feet Therefore, if a ball is thrown directly upward from an initial height of 200 feet with an initial velocity of 96 feet per second, after 3 seconds it will reach a maximum height of 344 feet.

Detailed solution to problem # We will begin by substituting our givens in to the projectile height formula: At time t = 0, v o = 96 ft/sec, and s o = 200 feet. The graph of the equation depicting the path of the ball is as follows: We want to know what the value of t will be when = 300. To find out, we substitute 300 for , and solve the quadratic equation for t. subtract 300 from each side of the equation solve for t using the quadratic formula For a quadratic in the form , the quadratic formula is stated as . We have obtained two values that represent the time that the ball reaches a height of 300 feet. The first value 1. 34 indicates that after 1. 34 seconds have passed, the ball is at a height of 300 feet. Then the ball reaches its maximum height and begins to fall back to the ground. After 4. 66 seconds it is once again at 300 feet. Then it will continue to fall to the ground. The answer we were seeking is 1. 34, the time the ball initially reached 300 feet after it has been thrown.