1. A ball rolls down a roof that makes an angle of 30° to the horizontal It rolls off the edge with a speed of 10.00 m/s. The distance to the ground from that point is 7.50 m. (a) How much time is the ball in the air? (b) How far from the base of the house does it land? (c) What is its speed just before landing?
Soluções para a tarefa
a) How much time is the ball in the air? 0,83 seconds
(b) How far from the base of the house does it land? 7,2m
(c) What is its speed just before landing? 11,9m/s
Uniformly Accelerated Rectilinear Motion
The ball is considered to have a motion similar to a projectile where the vertical motion determines the time in the air. Gravity is assumed to be 9.81m/s².
A velocity V0 = 10m/s is given, at an angle of 30°. Decomposing this velocity for the x and y axes, we have:
- Vertical component of velocity (Vy)
Vy = V0 * sen 30 = 10 * 0,5 → Vy = 5m/s
- Horizontal component of velocity (Vx)
Vx = V0 * cos 30 = 10 * √3/2 = 8,66m/s
It was also reported that the distance to the ground (h) = 7,5m.
(a) How much time is the ball in the air?
To calculate the time, the equation is used:
x = X0 + V0 * t + 1/2 at², making some modifications, one has:
x-x0 = V0 * t + 1/2 at².
x-x0 = The distance to the ground = Δy = 7,5m
V0 = vertical component of velocity = Vy = 5m/s
a = gravity = 9,8m/s²
Therefore:
7,5 = 5t + 1/2*9,8*t²
Multiply the whole equation by 2: 15 = 10t + 9,8t²
Then: 9,8t²+10t-15=0.
The second degree equation can be solved by Bhaskara:
Δ=b²-4ac = 10²-4*9,8*-15 = 100 + 588 = 688
t = -b ± √Δ/2A
t1 = -b -√Δ/2A and t2 = -b +√Δ/2A
t1 = -10 - 26,23 /2*9,8 = -36,23 / 19,6 = -1,85s
t2 = -10 + 26,23 /2*9,8 = 16,23 / 19,6 = 0,83s
There is no negative time, so the value of t1 is excluded. So the time the ball stayed in the air is 0,83s.
(b) How far from the base of the house does it land?
The distance from the base of the house is Δx. We use the same formula as in "a".
x-x0 = V0 * t + 1/2 at².
x-x0 = Δx
V0 = horizontal component of velocity = Vx = 8,66m/s
t = 0,83s;
a = 0 - Do not consider the horizontal gravity.
Then:
Δx = 8,66 * 0,83 + 1/2 * 0 * 0,83²
Δx = 8,66 * 0,83 + 0
Δx = 7,188 ≅ 7,2m
(c) What is its speed just before landing?
To calculate the velocity we use the formula:
v² = V0² + 2gd
v = fall velocity (m/s);
V0 = initial velocity = 10m/s
g = gravity = 9,8m/s²;
d = horizontal distance = 7,2m
Therefore:
V² = 10² + 2 * 9,8 * 7,2
V² = 100 + 141,2
v² = 241,2
V = 11,9 m /s
For better fixation of the content you can check another question about Uniformly Accelerated Rectilinear Motion at the link: https://brainly.com.br/tarefa/47686425
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