Question

In Sussex County, Delaware, a post-Halloween tradition is "Punkin Chunkin," in which contestants build cannons, catapults, trebuchets, and other devices to launch pumpkins and compete for the greatest distance. Though hard to believe, pumpkins have been projected a distance of 4086 feet in this contest. What is the minimum initial speed needed for a shot of 3954 ft?

Answer 1

General guidance

Concepts and reason

The concept required to solve the given problem is Projectile motion of a body.

The maximum range of the pumpkin is given; therefore, using the formula for maximum range of a projectile its initial speed can be determined.

Fundamentals

A projectile motion is a two-dimension motion. The analysis of a projectile motion is done by splitting it into two components, that is vertical and horizontal

Maximum height Range

The initial velocity V is split into two components

Along the horizontal-V cos \u03b8 Along the vertical-V sin \u03b8

In the vertical direction, the object will experience a constant acceleration due to gravity. Therefore,

ay -g

Here is the component of acceleration along the vertical and is the acceleration due to gravity

This acceleration will oppose the motion of the projectile as it goes up. Therefore, its vertical component of velocity will decrease with time and hence will become zero at the maximum height.

The time taken to by the object to reach its maximum height is

max height = usino

Here, is the initial velocity, is the angle with the horizontal at which projectile is thrown and is the acceleration due to gravity.

The taken by the object to fall back from the maximum height back to ground will be same.

Therefore, total time of flight is

2u sin \u03b8

As there is no external force in the horizontal direction, the acceleration along the horizontal direction is zero

ax

The distance the projectile travels along the horizontal direction is called Range.

Therefore,

R- horizontalspeedxtimeof flight (11 coso( 2"sino R R (14 cos \u03b8 112 sin 2\u03b8

Step-by-step

Step 1 of 2

The range of a projectile is given by

i/ sin 2\u03b8

Therefore, when the velocity is minimum the angle at which projectile should be thrown so that the desired range is achieved is when is maximum.

sin 2\u03b8

The maximum value of is one.

sin 2\u03b8

Thus,

sin 29-1 sin 2\u03b8-sin 900 20 90 \u03b8 450

Therefore, the range for minimum velocity is given by the following.

Explanation | Common mistakes | Hint for next step

The range of a projectile depends on the angle and magnitude of the initial velocity. To attain a certain range at minimum velocity, the angle of throw should be maximum. Hence, the angle will be 45⁰. By substituting the maximum value of in the formula of range, the expression for range at minimum velocity was derived.

sin 2\u03b8

Step 2 of 2

The range at which pumpkins are thrown =

R = 4086ft R=(4086ft)(0.3048 ) R = 1245.41 m

The expression for range attained at minimum velocity is given by the following.

Substituting
1245.41m
for and
9.80 s2
for

\u200e

245.41mmin 9.80 i12217.47 Umin -110.53

The minimum velocity is

The minimum velocity at which a certain range is achieved is derived in the previous part. The range at which pumpkins were thrown is given. The minimum velocity at which they were thrown to reach this distance was calculated by substituting the value of range achieved by the pumpkin and the value of acceleration due to gravity.

Answer

The minimum velocity is
Uin0.53 110,.3

Answer only

The minimum velocity is
Uin0.53 110,.3