Help! I've been trying to understand this question for quite some time now and every attempt i make at it i get the wrong answer. maybe i'm just pluging the numbers into the equations wrong, but i always get the wrong answers. if someone can just breakdown the steps for me it would be greatly appreciated.
A projectile is fired with an initial speed of 33.0 m/s at an angle of 15.0 degrees above the horizontal. The object hits the ground 8.0 sec later.
A) how much highter or lower is the launch point relative to the point where the projectile hits the ground?
B) to what maximum height above the launch point does the projectil rise?
C) what is the magnitude of the projectile's velocity at the instant it hits the ground?
D) what is the direction (below +x) of the projectile's velocity at the instant it hits the ground?
a wowen walk a distance of 225m with a aveage speed of 1.5m/s what time is required to walk.
To solve these questions, we can use the equations of motion for projectile motion. Let's break down each part step by step:
A) To find how much higher or lower the launch point is relative to the point where the projectile hits the ground, we need to determine the vertical displacement.
Step 1: Split the initial velocity into its vertical and horizontal components.
- The vertical component is given by V_y = V_initial * sin(theta), where V_initial is the initial speed (33.0 m/s) and theta is the launch angle (15.0 degrees).
- The horizontal component is given by V_x = V_initial * cos(theta).
Step 2: Calculate the time taken to reach the ground.
- The time of flight (t) is given in the question as 8.0 seconds.
Step 3: Calculate the vertical displacement (h) using the formula h = V_y * t + (1/2) * g * t^2.
- g is the acceleration due to gravity, approximately 9.8 m/s^2.
B) To find the maximum height above the launch point, we need to determine the vertical displacement at the highest point of the projectile's trajectory.
Step 1: Calculate the time taken to reach the maximum height.
- In projectile motion, the time taken to reach maximum height is equal to half of the total time of flight.
- So, t_max = t / 2 (where t is the time of flight mentioned in the question).
Step 2: Calculate the vertical displacement using the formula h_max = V_y * t_max + (1/2) * g * t_max^2.
C) To find the magnitude of the projectile's velocity at the instant it hits the ground, we need to determine the resultant velocity.
Step 1: Calculate the vertical component of the final velocity (V_y_f) at the instant it hits the ground.
- The final vertical velocity is given by V_y_f = V_y - g * t, where V_y is the vertical component of the initial velocity and t is the time of flight.
Step 2: Calculate the horizontal component of the final velocity (V_x_f).
- The horizontal component of the velocity remains constant throughout the motion, so V_x_f = V_x.
Step 3: Calculate the resultant velocity using the formula V_f = sqrt(V_x_f^2 + V_y_f^2).
D) To find the direction (below +x) of the projectile's velocity at the instant it hits the ground, we need to determine the angle at which the velocity vector points.
Step 1: Calculate the angle (theta_f) using the formula theta_f = atan(V_y_f / V_x_f).
- atan refers to the inverse tangent function.
Step 2: Convert the angle from radians to degrees if necessary.
By following these steps and plugging in the given values, you should be able to find the answers to the questions.