In an action movie the lead character must jump off a balcony at an angle of 20 degrees, 10 meters above the ground and land in a window 3 meters away and 8.5 metters abover the ground.
What should the velocity be in order to make this jump?
How long is the stunt double in the air?
What is the velocity as the stunt double impacts the window?
What is the maximum height the double is above the ground?
What is the acceleration at the highest point of the jump?
If a ball is thrown with a velocity of 25m/s at an angle of 37degrees above the horizontal, what is the vertical component of the veolocity
To solve these questions, we can analyze the motion of the stunt double using the principles of projectile motion. We'll assume that air resistance is negligible and use the constant acceleration due to gravity, which is approximately 9.8 m/s².
1. Firstly, let's find the initial velocity needed for the stunt double to jump off the balcony and land in the window. We can use the range equation for projectile motion:
Range = (Initial Velocity)² * sin(2θ) / acceleration due to gravity
Given:
Angle (θ) = 20 degrees = 20 * π / 180 radians
Range = 3 meters
Rearranging the equation:
Initial Velocity = √(Range * acceleration due to gravity / sin(2θ))
Plugging in the values:
Initial Velocity = √(3 * 9.8 / sin(2 * 20 * π / 180))
Solving this calculation will give us the answer to the first question.
2. To find the time the stunt double is in the air, we can use the vertical motion equation:
Vertical Displacement = Initial Velocityy * Time - (1/2) * acceleration due to gravity * Time²
Given:
Initial Vertical Displacement = 10 meters
Rearranging the equation:
Time = (Initial Velocityy ± √(Initial Velocityy² + 2 * acceleration due to gravity * Vertical Displacement)) / acceleration due to gravity
Note: We use the positive square root because we are interested in the time it takes to reach the highest point and then fall back down. The negative square root would give us the time for the stunt double to impact the ground.
Solve this equation to get the time the stunt double is in the air.
3. To find the velocity as the stunt double impacts the window, we can use the vertical motion equation again:
Velocity² = Initial Velocityy² + 2 * acceleration due to gravity * Vertical Displacement
Given:
Vertical Displacement = (8.5 - 10) meters (We need to account for the stunt double's higher position than the ground)
Solve this equation to find the velocity at the moment of impact.
4. The maximum height the stunt double reaches can be found using the equation for vertical displacement:
Vertical Displacement = Initial Velocityy * Time - (1/2) * acceleration due to gravity * Time²
The initial vertical velocity in this case is the component of the initial velocity perpendicular to the ground:
Initial Velocityy = Initial Velocity * sin(θ)
Plugging in the values, with the time found earlier, will give us the maximum height.
5. Finally, the acceleration at the highest point of the jump is equal to the acceleration due to gravity, since there are no other forces acting on the stunt double at that moment.
Therefore, the acceleration at the highest point is approximately 9.8 m/s².
By following these steps and plugging in the appropriate values, you can find the answers to each of the questions.