In an outdoor game a small bag of
rice is thrown from near the
ground with a velocity of 3.50 m/s
at an angle of 30.00 above the
horizontal. Find out how long the
bag will be in the air before it
lands at the same height from which it was thrown and how far it travelled horizontally.
time in air:
hf=hi+vo*sinTheta*t-1/2 g t^2
0=0+3.5*sin30*t-4.8t^2
t(4.8t-3.5*1/2)=0
solve for t.
time till velocity (v=v0-at) is zero:
3.5/2 - 9.8t = 0
t = 1.75/9.8
That's how long it rose upward. It fell for the same time, so ...
To find the time the bag will be in the air and the horizontal distance it will travel, we can use the equations of motion.
1. Find the time the bag will be in the air:
The bag will follow a projectile motion, moving both horizontally and vertically. We can analyze the vertical motion using the equation:
h = v₀y * t + (1/2) * g * t²
where:
- h is the vertical height (which is 0 since it lands at the same height as it is thrown),
- v₀y is the initial vertical velocity (v₀ * sinθ),
- t is the time the bag is in the air,
- g is the acceleration due to gravity (-9.8 m/s²).
Since the bag lands at the same height from which it was thrown, h = 0. We can rearrange the equation to solve for t:
0 = (v₀ * sinθ) * t + (1/2) * (-9.8) * t²
Simplifying the equation, we have:
0 = (3.50 * sin30°) * t - 4.9 * t²
0 = 1.75t - 4.9t²
Rearranging the equation, we get:
4.9t² - 1.75t = 0
Using the quadratic formula:
t = (-b ± sqrt(b² - 4ac)) / 2a
In this case, a = 4.9, b = -1.75, and c = 0.
Plugging in the values, we find:
t₁ = (-(-1.75) + sqrt((-1.75)² - 4(4.9)(0))) / (2 * 4.9)
= 0.39 seconds (approximately)
t₂ = (-(-1.75) - sqrt((-1.75)² - 4(4.9)(0))) / (2 * 4.9)
= 0.00 seconds (approximately)
Since time cannot be negative and we discard the zero value for t because it is not physically meaningful in this context, the bag will be in the air for approximately 0.39 seconds.
2. Find the horizontal distance traveled:
The horizontal distance can be found using the equation:
d = v₀x * t
where:
- d is the horizontal distance,
- v₀x is the initial horizontal velocity (v₀ * cosθ),
- t is the time the bag is in the air.
Plugging in the values:
d = (3.50 * cos30°) * 0.39
= 1.52 meters (approximately)
So, the bag will be in the air for approximately 0.39 seconds and travel a horizontal distance of approximately 1.52 meters.