Stuntman jumping 2D kinematics.

[CandyApples]

Homework Statement

A stuntman is jumping from building A to building B. His flight lasts 1s. He leaves point A with a speed of V_o at an angle of 30 degrees above horizontal. Building B is .9m shorter than building A. What horizontal distance was covered by the stuntman given he makes it to the very edge of building B? What was the maximum height with respect to the starting point attained by the stuntman?

Homework Equations

Displacement and velocity kinematic equations.

The Attempt at a Solution

displacement x= cos(30)*V_o*1s
displacement y= sin(30)*v_o+.5(-9.8)(1)-.9m

I know that somehow i need to find Vo from the Y data then plug it into the x displacement equation but i am not sure how this is going to be possible with the given information.

 

[method_man]

Homework Statement

A stuntman is jumping from building A to building B. His flight lasts 1s. He leaves point A with a speed of V_o at an angle of 30 degrees above horizontal. Building B is .9m shorter than building A. What horizontal distance was covered by the stuntman given he makes it to the very edge of building B? What was the maximum height with respect to the starting point attained by the stuntman?

Homework Equations

Displacement and velocity kinematic equations.

The Attempt at a Solution

displacement x= cos(30)*V_o*1s
displacement y= sin(30)*v_o+.5(-9.8)(1)-.9m

I know that somehow i need to find Vo from the Y data then plug it into the x displacement equation but i am not sure how this is going to be possible with the given information.

Consider building A as a xy origin, and building B as a final point. When you look at the formula for vertical displacement y=-1/2*g*t²+v_y*t, you know the time t and position y at that time t. Take under consideration that y is negative (-9).

 

[tomcue]

I would approach this problem by first defining the variables and identifying the known and unknown quantities. In this case, the known quantities are the flight time (1s), initial velocity (Vo), and angle of launch (30 degrees). The unknown quantities are the horizontal distance covered and the maximum height attained.

Next, I would use the kinematic equations to solve for the unknown quantities. The x-displacement equation is x = V0tcos(theta), where x is the horizontal distance, V0 is the initial velocity, t is the time, and theta is the angle of launch. Plugging in the known values, we get x = Vo(1)cos(30) = 0.866Vo.

To solve for Vo, we can use the y-displacement equation, y = V0tsin(theta) - 1/2gt^2, where y is the vertical displacement and g is the acceleration due to gravity (9.8 m/s^2). We can rearrange this equation to solve for Vo: Vo = y/(tsin(theta)) + 1/2gt. Plugging in the known values, we get Vo = -0.9/(1sin(30)) + 1/2(9.8)(1) = 2.1 m/s.

Now that we have Vo, we can plug it into the x-displacement equation to solve for the horizontal distance: x = (2.1 m/s)(1s)cos(30) = 1.825 m. Therefore, the stuntman covered a horizontal distance of 1.825 m to reach the edge of building B.

To find the maximum height attained, we can use the y-displacement equation again: y = V0tsin(theta) - 1/2gt^2. Plugging in the known values, we get y = (2.1 m/s)(1s)sin(30) - 1/2(9.8)(1)^2 = 0.55 m. Therefore, the stuntman reached a maximum height of 0.55 m above the starting point.

In conclusion, using the given information and the kinematic equations, we can calculate that the stuntman covered a horizontal distance of 1.825 m and reached a maximum height of 0.55 m above the starting point. These calculations assume ideal conditions and do not take into account air resistance or other factors that