Answer:
11.7 m/s
Explanation:
To find its speed, we first find the acceleration of the center of mass of a rolling object is given by
a = gsinθ/(1 + I/MR²) where θ = angle of slope = 4, I = moment of inertia of basketball = 2/3MR²
a = 9.8 m/s²sin4(1 + 2/3MR²/MR²)
= 9.8 m/s²sin4(1 + 2/3)
= 9.8 m/s²sin4 × (5/3)
= 1.14 m/s²
To find its speed v after rolling for 60 m, we use
v² = u² + 2as where u = initial speed = 0 (since it starts from rest), s = 60 m
v = √(u² + 2as) = √(0² + 2 × 1.14 m/s × 60 m) = √136.8 = 11.7 m/s
Answer:
The arrow is at a height of 500 feet at time t = 2.35 seconds.
Explanation:
It is given that,
An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s
The projectile formula is given by :
We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,
On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds
So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.