Instructions: The simulation below illustrates the motion of a projectile launch
ID: 1572681 • Letter: I
Question
Instructions: The simulation below illustrates the motion of a projectile launched at different speeds and angles.
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A projectile is launched with a launch angle of 65°with respect to the horizontal direction and with initial speed 20 m/s.
(A) How do the vertical and horizontal components of its velocity vary with time?
(B) How long does it remain in flight?
(C) For a given launch speed, what launch angle produces the longest time of flight?
Conceptualize
Consider the projectile to be a point mass that starts with an initial velocity, upward and to the right, with forces from air resistance neglected. There is no force acting horizontally to accelerate its horizontal motion, while its vertical motion is accelerated downward by gravity. Therefore as the projectile moves to the right at a constant rate, the vertical part of its motion consists of first rising upward and then moving downward until the projectile strikes the ground. Use the simulation to display the projectile motion.
Categorize
The velocity has components in both the x and y directions, so we categorize this as a problem involving particle motion in two dimensions. The particle also has only a y component of acceleration, so we categorize it as a particle under constant acceleration in the y direction and constant velocity in the x direction.
(A) How do the vertical and horizontal components of its velocity vary with time?
Analyze
The initial velocity in the x-direction vx0 is related to the initial speed by
vx0 = v0 cos(65°)
The constant velocity in the x-direction means that the equation describing the time dependence of x for the particle, with x0 taken as 0, is
x = x0 + vx0t = 0 +
m/s
t
The equation for the vertical coordinate, which is constantly accelerating downward at g = 9.8 m/s2, is
y = y0 + vy0t
gt2 =
m/s
t +
m/s2
t2
Finalize
The ½gt2 term is negative. The other time-dependent term is proportional to t and positive. Which of the two dominates at small t ? Which term's magnitude gets larger faster as t gets large? What effect does that have on the sign of the y coordinate as t starts out small and then gets larger? Is this consistent with the path you expect the projectile to take?
(B) How long does it remain in flight?
Analyze
The y-component of the projectile's velocity decreases by 9.8 m/s for each second of flight as the projectile rises. Therefore it takes a time of
ty,max =
=
for the vertical component of velocity to reach a value of 0, which occurs at the projectile's maximum height. At each height on the way down the particle has regained the same speed and has the same acceleration as it had on the way up, so that the complete time of flight is twice the time to reach the maximum height, and is equal to
tflight =
In the present problem, that expression gives
tflight = s
(C) For a given launch speed, what launch angle produces the longest time of flight?
Analyze
The time of flight for a given initial speed vi,
tflight =
is largest when sin is largest, which is at = °.
What does that correspond to physically?
Finalize
Is the result consistent with experience? Try the simulation for various launch angles and speeds, and check if the computer calculation shows this same result.
This electronic presentation to be used with SERWAY/JEWETT. Physics for Scientists and Engineers with Modern, Hybrid (with Enhanced WebAssign Homework and eBook LOE Printed Access Card for Multi Term Math and Science), 8E. From SERWAY/JEWETT.Physics for Scientists and Engineers with Modern, Hybrid (with Enhanced WebAssign Homework and eBook LOE Printed Access Card for Multi Term Math and Science), 8E. © 2012 Brooks/Cole, a part of Cengage Learning, Inc. All rights reserved. Reproduced by permission. Text/images may not be modified or reproduced in any way without prior written permission of the publisher. www.cengage.com/permissions
Projectile Motion
Explanation / Answer
Given,
v0 = 20 m/s ; theta = 65 deg
A)The horizontal veclotiy of the projectile remains same with time and through put the motion. The horizontal velocity is given by:
v0x = vx cos(theta)
But the vertical velocity keeps changing with time. at any istant of time its given by:
v0y' = v0y + g t
where, v0y = v0 sin(theta)
B)We know from eqn of motion
v = u + at
v = 0 at max height ; u = v0y and a = -g
u = gt => t = u/a
t = 20 sin65/9.8 = 1.85 s
T = 2t = 2 x 1.85 = 3.7 s
Hence, T = 3.7 s
C)The largest value of time will be given by:
T = 2 vo sin(theta)/g
sin(theta) is max for (theta) = 90 deg
Hence, theta = 90 deg