Read and analyze the situation below and solve for the unknown values. Follow the GRESA pattern in solving the problem.
G - given
R - required
E - equation
S - solution
A - answer
1. A student threw a ball horizontally out of a window 8.0 m above the ground. It was caught by another student who was 10.0 m away. What was the initial velocity of the ball?
2. A baseball was hit at 45 m/s at an angle of 45 degrees above the horizontal.
a. How long did it remain in the air?
b. How far did ut travel horizontally?
3. A bullet travelling 8.0 × 10² m/s horizontally hits a target 100 away. How far does the bullet fall before it hits the target?
G - given
R - required
E - equation
S - solution
A - answer
1. A student threw a ball horizontally out of a window 8.0 m above the ground. It was caught by another student who was 10.0 m away. What was the initial velocity of the ball?
2. A baseball was hit at 45 m/s at an angle of 45 degrees above the horizontal.
a. How long did it remain in the air?
b. How far did ut travel horizontally?
3. A bullet travelling 8.0 × 10² m/s horizontally hits a target 100 away. How far does the bullet fall before it hits the target?
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Step-by-step explanation:
1.)The horizontal distance traveled by a ball
d=v_0td=v
0
t
The time of a ball free falling
t=\sqrt{2h/g}t=
2h/g
So, the initial velocity of a ball
v_0=\frac{d}{t}=\frac{d}{\sqrt{2h/g}}v
0
=
t
d
=
2h/g
d
=\frac{10.0\:\rm m}{\sqrt{2\times 8.0\:\rm m/(9.8\:\rm m/s^2)}}=7.83\:\rm m/s=
2×8.0m/(9.8m/s
2
)
10.0m
=7.83m/s
3.)answer: 5/64m
the speed of the bullet : 8.0×10^2m/s
the time : t
100= VT=8.0×10^2.t
t=1/8 s
h= 1/2 gt^2(AS g= 10mls^2
= 1/2 ×10 ×(1/8) ^2
answer is : 5/64m ❤