Answer:
V = 48.4m/s, θ = 73.6° below the horizontal.
Explanation:
Given h = 109.53m, x = range = 65m,
θ = 0°
This problem involves the concepts of projectile motion.
Let yo = h = 109.53m
y = final position on the y axis. = 0m (ground level)
y = yo + vosinθ – 1/2gt²
0 = 109.53 + vosin0° – 1/2×9.8×t²
0 = 109.53 – 4.9t²
4.9t² = 109.53
t² = 109.53/4.9 = 22.35
t = √22.35 = 4.73s
So it takes the ball t = 4.73 seconds to get to the ground from the launch point.
x = (vocosθ)×t
65 = (vocos0°)×4.73
65/4.73 = vo
Vo = 13.7m/s
Vx = Vox = Vocosθ = 13.7cos0° = 13.7m/s
Vy = Voy – gt = Vosinθ – gt
Vy = 13.7sin0° – 9.8×4.73 = –46.4m/s
V = √(Vy² + Vx²) = √(-46.4² + 13.7²)
V = 48.4m/s
θ = Tan-¹(vy/vx) = tan-¹(-46.4/13.7) = -73.6°
θ = 73.6° below the horizontal
V = 48.4m/s
Recall the formula,
∆<em>θ</em> = <em>ω</em>₀ <em>t</em> + 1/2 <em>α</em> <em>t</em> ²
where ∆<em>θ</em> = angular displacement, <em>ω</em>₀ = initial angular speed (which is zero because the disk starts at rest), <em>α</em> = angular acceleration, and <em>t</em> = time. Solve for the acceleration with the given information:
50 rad = 1/2 <em>α</em> (5 s)²
<em>α</em> = (100 rad) / (25 s²)
<em>α</em> = 4 rad/s²
Now find the angular speed <em>ω </em>after 3 s using the formula,
<em>ω</em> = <em>ω</em>₀ + <em>α</em> <em>t</em>
<em>ω</em> = (4 rad/s²) (3 s)
<em>ω</em> = 12 rad/s
Answer:
Option b
Explanation:
An object is said to fall freely when there is no force acting on the object other than the gravitational force. Thus the acceleration of the object is solely due to gravity and no other acceleration acts on the body.
Also the initial velocity of the body in free fall is zero and hence less than the final velocity.
As the body falls down closer to earth, it experiences more gravitational pull and the velocity increases as it falls down and the moment it touches the ground the period of free fall ends at that instant.
Thus the final velocity of an object in free fall is not zero because the final velocity is the velocity before coming in contact with the ground.