1.Solve sin ¥è = 0.95 for -90¨¬ ¡ ¥è ¡ 90¨¬.
2.A volleyball player spikes the ball from a height of 2.44 meters. Assume that the path of the ball is a straight line. To the nearest degree, what is the maximum angle, ¥è, at which the ball can be hit and land within the court? (Enter only the number.)
If you could give me steps that are pretty simple to follow to get the correct answer that would be great. I'm just really confused. Thank You.
#1. no calculator? It will be in 1st quadrant.
#2. how long is the court? The angle is x where tan(x) = 2.44/(court length)
1. To solve sin ¥è = 0.95, we need to find the values of ¥è between -90¨¬ and 90¨¬ that satisfy the equation.
First, let's find the angle ¥è in the first quadrant (0¨¬ ≤ ¥è ≤ 90¨¬). We can use the inverse sine function (also known as arcsine) to find the angle with a given sine value.
Step 1: Take the inverse sine (arcsine) of 0.95.
¥è₁ = arcsin(0.95)
Step 2: Use a calculator to find the value of ¥è₁.
Now let's find the angle ¥è in the second quadrant (90¨¬ ≤ ¥è ≤ 180¨¬). Since sine is positive in the second quadrant, we consider the supplementary angle to the angle found in step 2.
Step 3: Subtract the value of ¥è₁ from 180¨¬ to find ¥è in the second quadrant.
¥è₂ = 180¨¬ - ¥è₁
Finally, we have the solutions for sin ¥è = 0.95 in the range -90¨¬ ≤ ¥è ≤ 90¨¬: ¥è₁ and ¥è₂. You can substitute the values of ¥è₁ and ¥è₂ into the equation to check if they indeed satisfy sin ¥è = 0.95.
2. To find the maximum angle ¥è at which the ball can be hit and land within the court, we can use trigonometry and apply the concept of the range of a projectile.
The range of a projectile, in this case, is the horizontal distance traveled by the ball after being spiked. We need to find the angle at which this range is maximized.
Step 1: Calculate the range of the projectile using the formula:
Range = (initial velocity squared) * sin(2¥è) / g
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Step 2: In this problem, the height from which the ball is spiked, 2.44 meters, represents the maximum height of the projectile. Use this maximum height and the range formula to calculate the initial velocity of the ball.
Step 3: Now that we have the initial velocity, we can find the angle ¥è that gives the maximum range. This is done by finding the maximum value of the function sin(2¥è). The maximum value of sin(2¥è) is 1, which occurs when 2¥è = 90¨¬. So we solve for ¥è as follows:
2¥è = 90¨¬
¥è = 90¨¬ / 2
¥è = 45¨¬
Therefore, to the nearest degree, the maximum angle ¥è at which the ball can be hit and land within the court is 45¨¬.