Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A hunter aims directly at a target (on the same level) 38.0 m away. (a) If the arrow leaves the bow at a speed of 23.1 m/s, by how much will it miss the target? (b) At what angle should the bow be aimed so the target will be hit?

Short Answer

Expert verified

The obtained values of the distance and angle are y=13.27mand θ=20.05°, respectively.

Step by step solution

01

Step 1. Calculate the time of flight

In this problem, consider the downward direction as positive y and the point at which the arrow leaves the bow as the origin. In order to determine the angle of launch, use the relation of level horizontal range.

Given data:

The hunter aims at the target at a distance of d=38m.

The speed of the arrow is v=23.1m/s.

The formula to calculate the time of flight can be written as:

t=dv

On plugging the values in the above relation, you get:

t=38m23.1m/st=1.645s

02

Step 2. Calculate the distance by which the hunter will miss the target

The formula to calculate the distance can be written as:

y=ut+12gt2.

Here, u is the velocity in the vertical direction whose value is zero, y is the distance by which the hunter misses the target, and g is the gravitational acceleration.

On plugging the values in the above relation, you get:

y=0t+129.81m/s21.645s2y=13.27m

Thus, y=13.27mis the distance by which the hunter misses the target.

03

Step 3. Calculate the launch angle

The relation to calculate the launch angle can be written as:

d=v2sin2θg

Here, θis the required angle.

On plugging the values in the above relation, you get:

38m=23.1m/s2sin2θ9.81m/s2θ=20.05°

Thus, θ=20.05°is the required launch angle.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Apollo astronauts took a ‘nine iron’ to the moon and hit a golf ball about 180 m. Assuming that the swing launch angle and so on were the same as on the earth, where the same astronauts could hit it only 32 m, estimate the acceleration due to gravity on the surface of the moon. (Neglect the air resistance in both cases, but on the moon, there is none.)

Suppose the rescue plane of Problem 31 releases the supplies a horizontal distance of 425 m in advance of the mountain climbers. What vertical velocity (up or down) should the supplies be given so that they arrive precisely at the climbers’ position (Fig. 3–39)? With what speed do the supplies land?

FIGURE 3-39Problem 32

Romeo is throwing pebbles gently up to Juliet’s window, and he wants the pebbles to hit the window with only a horizontal component of velocity. He is standing at the edge of a rose garden 8.0 m below her window and 8.5 m from the base of the wall (Fig. 3-49). How fast are the pebbles going when they hit her window?

Estimate by what factor a person can jump farther on the Moon as compared to the Earth if the take-off speed and angle are the same. The acceleration due to gravity on the Moon is one-sixth what it is on Earth.

Vis a vector 24.8 units in magnitude and points at an angle of 23.4oabove the negative X-axis. (a) Sketch this vector. (b) CalculateVxandVy. UseVxandVyobtain (again) the magnitude and direction ofV. [Note: Part (c) is a good way to check if you’ve resolved your vector correctly.]

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free