Ch 3.3 Help
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Ch 3.3 Help
Question 1: (1 point)
A football is kicked from level ground and lands after 1.076 s back on the ground. What is the vertical component of the velocity with which the ball leaves the kicker's foot (sheet 17,19,21)?
How to solve:
first divide t by 2 to get the time of the maximum height where V=0
t/2=.538
use equation V=V0+At to solve for V0
0=V09.81*.538
V0=9.81*.538
V0=5.2778
Question 2: (1 point)
A projectile is launched horizontally from a height of 1.538 m and hits the ground at the horizontal distance of 3.195 m from the launch point. What is the launch velocity ( see sheet 17,19, 25')?
How to solve:
Use equation Y=1/2gt^2 to solve for t
1.538=1/2*9.81*t^2
t^2=.3136
t=.55996
then plug into v=d/t to solve for v
v=3.195/.55996
v=5.7057
Question 3: (1 point)
A marble is dropped at the same time when a bullet is fired horizontally (both from the same height). The marble hits the ground after 0.6949 s. When does the bullet hit the ground (recall: no air resistance, see sheet 25,25')?
How to solve:
The bullet would hit the ground at the same time as the marble because gravity is the same for both.
t=.6949
Question 4: (1 point)
The projectile from a bebegun which is pointed vertically up reaches a maximum height h1 (see sketch Ch2 sheet 21). The maximum height when the same gun is tilted (sheet 21) at an angle of 29.39 degrees with respect to the horizontal is h2. What is the ratio h2/h1 (sheet 17,19)?
How to solve:
Take the sin of h2 and square it
sin(29.39)=.49075
.49075^2=.2408
A football is kicked from level ground and lands after 1.076 s back on the ground. What is the vertical component of the velocity with which the ball leaves the kicker's foot (sheet 17,19,21)?
How to solve:
first divide t by 2 to get the time of the maximum height where V=0
t/2=.538
use equation V=V0+At to solve for V0
0=V09.81*.538
V0=9.81*.538
V0=5.2778
Question 2: (1 point)
A projectile is launched horizontally from a height of 1.538 m and hits the ground at the horizontal distance of 3.195 m from the launch point. What is the launch velocity ( see sheet 17,19, 25')?
How to solve:
Use equation Y=1/2gt^2 to solve for t
1.538=1/2*9.81*t^2
t^2=.3136
t=.55996
then plug into v=d/t to solve for v
v=3.195/.55996
v=5.7057
Question 3: (1 point)
A marble is dropped at the same time when a bullet is fired horizontally (both from the same height). The marble hits the ground after 0.6949 s. When does the bullet hit the ground (recall: no air resistance, see sheet 25,25')?
How to solve:
The bullet would hit the ground at the same time as the marble because gravity is the same for both.
t=.6949
Question 4: (1 point)
The projectile from a bebegun which is pointed vertically up reaches a maximum height h1 (see sketch Ch2 sheet 21). The maximum height when the same gun is tilted (sheet 21) at an angle of 29.39 degrees with respect to the horizontal is h2. What is the ratio h2/h1 (sheet 17,19)?
How to solve:
Take the sin of h2 and square it
sin(29.39)=.49075
.49075^2=.2408
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