C is the answer to the question
The velocity at the maximum height will always be 0. Therefore, you will count your final velocity as 0, and your initial velocity as 35 m/s. Next, we know that the acceleration will be 9.8 m/s^2. How? Because the ball is thrown directly upward, and the only force acting on it will be the force of gravity pushing it back down.
The formula we use is h = (Vf^2 - Vi^2) / (2*-9.8m/s^2)
Plugging everything in, we have h = (0-1225)/(19.6) = 62.5 meters is the maximum height.
Ahhh this going to be confusing sorry...
1. α = Δω / Δt = 28 rad/s / 19s = 1.47 rad/s²
2. Θ = ½αt² = ½ * 1.47rad/s² * (19s)² = 266 rads
3. I = ½mr² = ½ * 8.7kg * (0.33m)² = 0.47 kg·m²
4. ΔEk = ½Iω² = ½ * 0.47kg·m² * (28rad/s)² = 186 J
5. a = α r = 1.47rad/s² * 0.33m = 0.49 m/s²
6. a = ω² r = (14rad/s)² * 0.33m = 65 m/s²
7. v = ω r = 28rad/s * ½(0.33m) = 4.62 m/s
8. s = Θ r = 266 rads * 0.33m = 88 m
If it takes
seconds to reach the car, then the distance
is
.
The bear's distance from the tourist's starting point is
For maximum
, we set the equations equal to each other:
so the distance is
Explanation:
We want to find the statement that is proven by the fact that the balls reach the same height.
A isn't supported by the evidence. Balls can reach the same height without having the same initial speed.
B isn't supported by the evidence. Balls can reach the same height without having the same launch angle.
C is supported. Projectiles spend the same amount of time going up as they do coming down, so if two projectiles reach the same height, then they must spend the same amount of time in the air.
D isn't supported by the evidence. Balls thrown at the same speed and complementary angles have the same range but different heights.
E isn't supported by the evidence. The mass of the ball doesn't affect the height.
