The method of heat transfer by which the sun warms the earth is
2 answers:
You might be interested in
Answer:
1) The maximum height reached is approximately 39.715 m
2) The time it takes the ball to reach the its highest point is approximately 2.85 s
Explanation:
1) The vertical velocity of the ball, u = 27.9 m/s
The acceleration due gravity, g = 9.8 m/s²
The kinematic equation that gives the height, h, reached by the ball is given as follows;
v² = u² - 2·g·h
Where;
v = The final velocity of the ball = 0 m/s at the maximum height
h = The height reached by the ball = at maximum height
Therefore, by substitution of the known values, we have;
v² = u² - 2·g·h
0² = 27.9² - 2 × 9.8 ×
2 × 9.8 × = 27.9²
= 27.9²/(2×9.8) ≈ 39.715
The maximum height reached = ≈ 39.715 m
2) From the kinematic equation of motion, v = u - g·t, we have the time, t, it takes the ball to reach the maximum height given as follows
At maximum height, the final velocity, v = 0 m/s, therefore, we have;
0 = 27.9 - 9.8 × t
9.8 × t = 27.9
t = 27.9/9.8 ≈ 2.85
The time it takes the ball to reach the maximum height = t ≈ 2.85 seconds.
Consider the projectile launched at initial velocity V at angle θ relative to the horizontal.
Neglect wind or aerodynamic resistance.
The initial vertical velocity is Vsinθ.
When the projectile reaches its maximum height of h, its vertical velocity will be zero.
If the time taken to attain maximum height is t, then
0 = Vsinθ - gt
t = (Vsinθ)/g, where g = acceleration due to gravity.
The horizontal component of launch velocity is Vcosθ. This velocity remains constant because aerodynamic resistance is ignored.
The time to travel the horizontal distance D is twice the value of t.
Therefore
D = Vcosθ*[(2Vsinθ)/g]
= (2V²sinθ cosθ)/g
= (V²sin2θ)/g
In order for D (horizontal distance) to be maximum,
That is,
Because
, therefore cos(2θ) = 0.
This is true when 2θ = π/2 => θ = π/4.
It has been shown that the maximum horizontal traveled can be attained when the launch angle is π/4 radians, or 45°.
Neglect wind or aerodynamic resistance.
The initial vertical velocity is Vsinθ.
When the projectile reaches its maximum height of h, its vertical velocity will be zero.
If the time taken to attain maximum height is t, then
0 = Vsinθ - gt
t = (Vsinθ)/g, where g = acceleration due to gravity.
The horizontal component of launch velocity is Vcosθ. This velocity remains constant because aerodynamic resistance is ignored.
The time to travel the horizontal distance D is twice the value of t.
Therefore
D = Vcosθ*[(2Vsinθ)/g]
= (2V²sinθ cosθ)/g
= (V²sin2θ)/g
In order for D (horizontal distance) to be maximum,
That is,
Because
This is true when 2θ = π/2 => θ = π/4.
It has been shown that the maximum horizontal traveled can be attained when the launch angle is π/4 radians, or 45°.