a horse gallops a distance of 60 meters in 15 seconds. then, he stops to eat some grass for 20 seconds. next, he trots for 25 se
conds over 60 meters. (the horse trots in the same direction he galloped.) finally, the horse races home (back to its starting position), traveling 120 meters in 20 seconds. graph the horse’s movements on the following distance-time chart.
1 answer:
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Answer:
1)
2)
Explanation:
<u>Projectile Motion</u>
When an object is launched near the Earth's surface forming an angle with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.
The heigh of an object can be computed as
Where is the initial height above the ground level,
is the vertical component of the initial velocity and t is the time
The y-component of the speed is
1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of
The object will reach the maximum height when . It allows us to compute the time to reach that point
Solving for
Thus, the maximum heigh is
We know this value is 8 meters
Solving for
Replacing the known values
2) We know at t=1.505 sec the ball is above Julie's head, we can compute
You will have to use this formula:
Final Velocity (V) = 4m/s
Initial Velocity (Vo) = 8m/s
Acceleration (a) = ? m/s^2
Time (t) = 2 secs
Then:
-> 4 = 8 + a x 2
-> 4 - 8 = 2a
-> -4 = 2a
-> a = -4/2
-> a = -2 m/s^2
Ps: It's value is negative because the she was in retrograde motion.
Answer: Her acceleration is -2 m/s^2.
Final Velocity (V) = 4m/s
Initial Velocity (Vo) = 8m/s
Acceleration (a) = ? m/s^2
Time (t) = 2 secs
Then:
-> 4 = 8 + a x 2
-> 4 - 8 = 2a
-> -4 = 2a
-> a = -4/2
-> a = -2 m/s^2
Ps: It's value is negative because the she was in retrograde motion.
Answer: Her acceleration is -2 m/s^2.