A golfer hits a ball from the origin with an initial speed of 35 m/s at an angle of 44 degrees above the horizontal axis. The ball is moving down when it lands on a green that is 3.00 m above the level where it was struck. Neglect air resistance.
1. How long is the ball in the air ? I know that the answer is "between 3.0 and 5.0 sec" but I don't know how this answer was achieved. How do I solve to get it?
2. How far as the ball traveled in the horizontal direction when it lands? The answer is "between 100 and 150 m" but again I don't know how to solve to get the answer.
3. At the highest point in the path, the ball's speed is:_____ The answer is "between 15 and 30 m/s"
4. What is the speed of the ball 1.8 seconds after it has been hit? The answer is "more than 25 m/s"
WHERE WAS THIS QUESTION FROM?
To solve these questions, we can use the equations of motion for projectile motion. In this case, the motion of the ball can be split into horizontal and vertical components.
1. To find the time of flight, we need to determine how long the ball stays in the air. We can use the vertical motion equation:
h = ut + (1/2)gt^2
where h is the vertical displacement (3.00 m), u is the initial vertical velocity (35 m/s × sin(44°)), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time of flight.
Solving for t, we have:
3.00 = (35 × sin(44°))t - 4.9t^2
-4.9t^2 + (35 × sin(44°))t - 3.00 = 0
By solving this quadratic equation, you will find two values of t. The time of flight will be positive, so discard the negative value. The time of flight will fall between these two values.
2. To find the horizontal distance traveled, we can use the horizontal motion equation:
s = ut
where s is the horizontal distance, u is the initial horizontal velocity (35 m/s × cos(44°)), and t is the time of flight obtained in the previous step.
Plugging in the values:
s = (35 × cos(44°))t
Calculate the value of s using the time of flight you obtained in step 1. The horizontal distance traveled will fall between these two values.
3. At the highest point in the path, the vertical velocity will be zero. To find the speed at this point, we can use the equation:
v = u + gt
where v is the final vertical velocity (0 m/s) and u is the initial vertical velocity (35 m/s × sin(44°)).
Solving for t, we have:
0 = (35 × sin(44°)) - 9.8t
Solve for t, and then use this value to calculate the magnitude of the velocity at that highest point. The speed will fall between these two values.
4. To find the speed of the ball 1.8 seconds after it has been hit, we can use the equation of motion:
v = u + gt
where v is the final velocity, u is the initial velocity (35 m/s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time (1.8 seconds).
Plugging in the values:
v = 35 + (-9.8 × 1.8)
Calculate the value of v. The speed of the ball will be more than this value.