Answer: 43 cm = 0.43 meters
5000, would be your answer.
Answer:
The ball shall keep rising tills its velocity becomes zero. Let it rise to a height h feet from point of projection.
Step-by-step explanation:
Let us take the point of projection of the ball as origin of the coordinate system, the upward direction as positive and down direction as negative.
Initial velocity u with which the ball is projected upwards = + 120 ft/s
Uniform acceleration a acting on the ball is to acceleration due to gravity = - 32 ft/s²
The ball shall keep rising tills its velocity becomes zero. Let it rise to a height h feet from point of projection.
Using the formula:
v² - u² = 2 a h,
where
u = initial velocity of the ball = +120 ft/s
v = final velocity of the ball at the highest point = 0 ft/s
a = uniform acceleration acting on the ball = -32 ft/s²
h = height attained
Substituting the values we get;
0² - 120² = 2 × (- 32) h
=> h = 120²/2 × 32 = 225 feet
The height of the ball from the ground at its highest point = 225 feet + 12 feet = 237 feet.
Answer:
32°
Step-by-step explanation:
Given:
∠DMQ = 58º
In this circle, the radius is DM. Since AD is tangent to the circle M, at point D, and the angle between a tangent and a radius is 90°
Therefore, ∠MDQ = 90°
The total angle in a triangle is 180°. Since we have the values of ∠MDQ and ∠DMQ, ∠DQM will be calculated as:
180 = ∠DMQ + ∠MDQ + ∠DQM
Solving for ∠DQM, we have:
∠DQM = 180 - ∠DMQ - ∠MDQ
∠DQM = 180 - 90 - 58
∠DQM = 32°
The measure of ∠DQM is 32°
