A ball is thrown from the top of a building upward at an angle of 66 ◦ to the horizontal and with an initial speed of 16 m/s. The ball is thrown at a height of 53 m above the ground and hits the ground 33.2074 m from the base of the building. The acceleration of gravity is 9.8 m/s2 . What is the speed of the ball just before it strikes the ground? Answer in units of m/s.

break up the initial velocity into vertical and horizontal components.

distancehorizontal=vix*t=16cos66*t
you know distance (33m),solve for time.

Vvf=Viv+gt Viv= 16sin66, and you know time.

vf=sqrt(vvf^2+vix^2)

To find the speed of the ball just before it strikes the ground, we need to use the equations of motion for projectile motion. Here are the steps to solve the problem:

Step 1: Decompose the initial velocity into its horizontal and vertical components.
The given initial speed of 16 m/s can be divided into horizontal and vertical components using trigonometry.
The horizontal component (Vx) remains constant throughout the motion and can be found using the equation Vx = V0 * cos(θ), where V0 is the initial speed, and θ is the angle with respect to the horizontal.
Vx = 16 m/s * cos(66°) ≈ 5.59 m/s.

Step 2: Find the time of flight for the ball.
We can find the time of flight for the ball by using the equation for vertical motion: h = V0y * t + (1/2) * g * t^2, where h is the vertical displacement, V0y is the vertical component of the initial speed, g is the acceleration due to gravity, and t is the time of flight.
In this problem, the vertical displacement is -53 m (negative because the ball is thrown upwards), V0y is the vertical component of the initial speed, which is V0y = V0 * sin(θ), and g is 9.8 m/s^2.
Using these values, the equation becomes: -53 m = (16 m/s * sin(66°)) * t + (1/2) * (9.8 m/s^2) * t^2.
Solving this quadratic equation will give two values of t. Since we are only interested in the positive value for time, we discard the negative value.

Step 3: Calculate the horizontal displacement of the ball.
The horizontal displacement can be found using the equation for horizontal motion: x = Vx * t, where x is the horizontal displacement and Vx is the horizontal component of the initial velocity.

Step 4: Calculate the speed just before the ball strikes the ground.
The final vertical velocity just before the ball hits the ground can be found using the equation: v = V0y + g * t, where v is the final vertical velocity, V0y is the vertical component of the initial velocity, g is the acceleration due to gravity, and t is the time of flight.

Step 5: Calculate the overall speed.
The overall speed of the ball just before it strikes the ground can be calculated using the Pythagorean theorem: speed^2 = (vx)^2 + (v)^2, where (vx) is the horizontal component of the velocity, and (v) is the vertical component of the velocity.