SOLUTION: the acceleration due to gravity on earth is 9.8m/s sqr a ball is thrown upward at an initial velocity of 15 m/s from a height of 1m off the ground
A) write a equation for the h
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Question 1066132: the acceleration due to gravity on earth is 9.8m/s sqr a ball is thrown upward at an initial velocity of 15 m/s from a height of 1m off the ground
A) write a equation for the height of the ball
B) what is the height after 1s
C) after how many seconds dose the ball hit the ground
D) what and when dose maximum height occur
E) repeat A&B for but g= 1.62m/s sqr
F) repeat A&B for but g= 23.1m/s sqr
Found 2 solutions by stanbon, ikleyn:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
the acceleration due to gravity on earth is 9.8m/s sqr a ball is thrown upward at an initial velocity of 15 m/s from a height of 1m off the ground
A) write a equation for the height of the ball(
h(t) = -9.8t^2+15t+1
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B) what is the height after 1s
h(1) = -9.8+15+1 = 6.2 meters
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C) after how many seconds does the ball hit the ground
Solve:: -9.8t^2+15t+1 = 0
t = 1.595 seconds
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D) what and when does maximum height occur?
Vertex occurs when t = -b/(2a) = -15/(2*-9.8) = 0.7653 seconds
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I'll leave the following to you.
Cheers,
Stan H.
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E) repeat A&B for but g= 1.62m/s sqr
F) repeat A&B for but g= 23.1m/s sqr
Answer by ikleyn(52778) (Show Source): You can put this solution on YOUR website!
.
In this site, there is a bunch of lessons on a projectile thrown/shot/launched vertically up
- Problem on a projectile moving vertically up and down
- Problem on an arrow shot vertically upward
- Problem on a ball thrown vertically up from the top of a tower
- Problem on a toy rocket launched vertically up from a tall platform
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this textbook under the topic "Projectiles launched/thrown and moving vertically up and dawn".
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