Call the starting point (xo, yo) = (0,0). To find (x, y) for the ending point, use triangle trig on the hill angled at 20°: x = d cos 20° = .940d and y = d sin 20° = .342d Plug in: .940d (vo cos eo) t (Eqn. 1) .342d = (vo sin 0o) t-gt² (Eqn. 2) The variable t is not mentioned in the problem. We can remove it by solving Eqn. 1 for t and plugging into Eqn. 2: t = .940d vo cos 80 .940d vo cos bo Cleaning up and plugging in g = 9.8: .342d = (vo sin o) .940d -) - 119 (20 COS 60 5-)² .342d.940d tan 00- (4.33) Divide equation by d (to cancel) and clean up some more: (4.33) d = (.940 tan 60 - .342) (vo cos 0o)² 10 cos fo Text 8 vo 84.0 Type your answer... vo m Type your answer... во 44.1° во 39.1⁰ d d 406m DO

Call the starting point (xo, yo) = (0,0). To find (x, y) for the ending point, use triangle trig on the hill angled at 20°: x = d cos 20° = .940d and y = d sin 20° = .342d Plug in: .940d (vo cos eo) t (Eqn. 1) .342d = (vo sin 0o) t-gt² (Eqn. 2) The variable t is not mentioned in the problem. We can remove it by solving Eqn. 1 for t and plugging into Eqn. 2: t = .940d vo cos 80 .940d vo cos bo Cleaning up and plugging in g = 9.8: .342d = (vo sin o) .940d -) - 119 (20 COS 60 5-)² .342d.940d tan 00- (4.33) Divide equation by d (to cancel) and clean up some more: (4.33) d = (.940 tan 60 - .342) (vo cos 0o)² 10 cos fo Text 8 vo 84.0 Type your answer... vo m Type your answer... во 44.1° во 39.1⁰ d d 406m DO

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter2: Vectors

Section: Chapter Questions

Problem 63P: Assuming the +x-axis is horizontal to the right for the vectors in the preceding figure, find (a)...

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Concept explainers

Scalars and Vectors

Scalars and Vectors are the classification of physical quantities that are used to denote various terms in physics.

Question

Solution:
The projectile motion equations:
x = x₁ + (vo cos 0) t
y = yo + (vo sin 0o) t - 1/2gt²
Call the starting point (xo, yo) = (0,0). To find (x, y) for the ending
point, use triangle trig on the hill angled at 20°:
x = d cos 20° = .940d and y = d sin 20° = .342d
Plug in:
.940d =
(vo coso) t (Eqn. 1)
.342d = (vo sin 0o) tgt² (Eqn. 2)
The variable t is not mentioned in the problem. We can remove it by solving
Eqn. 1 for t and plugging into Eqn. 2:
t
=
.940d
vo cos 00
.940d
vo cos o
-) - 12/19 (
Cleaning up and plugging in g = 9.8m:
.342d = (v₁ sino) (
.940d
vo cos 00
2

Call the starting point (xo, Yo) = (0, 0). To find (x, y) for the ending
point, use triangle trig on the hill angled at 20°:
X = d cos 20° =
.940d and y = d sin 20° = .342d
Plug in:
.940d = (vo cos
) t (Eqn. 1)
.342d = (vo sin 0o) t-gt² (Eqn. 2)
The variable t is not mentioned in the problem. We can remove it by
solving Eqn. 1 for t and plugging into Eqn. 2:
t
.940d
vo cos 00
.342d
=
(vo sin o)
.940d
vo cos 00
-910
Cleaning up and plugging in g
-
9.8:
.342d = .940d tan 0o – (4.33™) (
.940d
vo cos o
2
0)²
d
vo cos o
2
Divide equation by d (to cancel) and clean up some more:
(4.33 ) d = (.940 tan o .342) (vo cos 0₁)²
Text
7
8
Vo
84.0m
S
Type your answer...
vo
m
S
Type your answer...
00
44.1⁰
00
39.1⁰
d
m
d
406m
D

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