Manny is playing golf with his friend, he hits the golf ball from a 7.0m high building with a velocity of 8.0 m/s [45° above the horizontal], and it lands next to a pond. How long does it take for the golf ball to hit the ground and reach the pond? vix viy 8.0m/s) cos 45 cos 45 = vix

Glencoe Physics: Principles and Problems, Student Edition
1st Edition
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Paul W. Zitzewitz
Chapter3: Accelerated Motion
Section3.2: Motion With Constant Acceleration
Problem 16PP
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write me one question on projectile motion in the forum with answer solutions
the question and the answer below is an examples for you to know what I am expecting you to answer don't copy the answer
below make a question similar to it and solve that
Manny is playing golf with his friend, he hits the golf ball from a 7.0m high
building with a velocity of 8.0 m/s [45° above the horizontal], and it lands next to
a pond. How long does it take for the golf ball to hit the ground and reach the
pond?
vix
viy
vix = (8.0m/s) cos 45
(right)
cos 45 =
vix = 5.65m/s
sin 45 = viy
viy - (8.0m/s) sin 45
viy = 5.65m/s (up)
a = 9.81m/s²
viy = 5.65m/s (up)
Ad 7.0m (down
(down)
vfy² = viy² + 2a A dy
vfy² = ((5.
5.65m/s) ²) + 2(−9.81m/s²) (-7.0m
vfy=√169.2625
vfy = ±13m/s
vfy = -13m/s
vf=vi+a At
– 13m/s = 5.65m/s + (−9.81m/s²) A t
At=
At = 1.9s
-18.65m/s
-9.81m/s²
Transcribed Image Text:write me one question on projectile motion in the forum with answer solutions the question and the answer below is an examples for you to know what I am expecting you to answer don't copy the answer below make a question similar to it and solve that Manny is playing golf with his friend, he hits the golf ball from a 7.0m high building with a velocity of 8.0 m/s [45° above the horizontal], and it lands next to a pond. How long does it take for the golf ball to hit the ground and reach the pond? vix viy vix = (8.0m/s) cos 45 (right) cos 45 = vix = 5.65m/s sin 45 = viy viy - (8.0m/s) sin 45 viy = 5.65m/s (up) a = 9.81m/s² viy = 5.65m/s (up) Ad 7.0m (down (down) vfy² = viy² + 2a A dy vfy² = ((5. 5.65m/s) ²) + 2(−9.81m/s²) (-7.0m vfy=√169.2625 vfy = ±13m/s vfy = -13m/s vf=vi+a At – 13m/s = 5.65m/s + (−9.81m/s²) A t At= At = 1.9s -18.65m/s -9.81m/s²
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