A ball is thrown with a speed of 19m/s and at an angle of 46 degrees. How high does it go above the cliff to the nearest tenth?
To find how high the ball goes above the cliff, you can analyze the motion of the ball in two separate directions: horizontal and vertical. Let's break it down step by step:
Step 1: Analyzing the horizontal motion
The initial horizontal velocity of the ball remains constant throughout its motion. Since there are no horizontal forces acting on the ball, the horizontal velocity will always be the same. Therefore, we can use the equation:
Horizontal distance = Horizontal velocity × Time
Since we are interested in the vertical motion, we don't need the exact horizontal distance for this particular question.
Step 2: Analyzing the vertical motion
The initial vertical velocity (Vo) can be found by breaking down the given velocity into its vertical and horizontal components. The vertical velocity is given by:
Vertical velocity (Vo) = Initial velocity × sin(angle)
Vo = 19 m/s × sin(46°)
Vo = 19 m/s × 0.7193
Vo ≈ 13.6687 m/s
Now that we have the initial vertical velocity, we can analyze the vertical motion using the following kinematic equation:
Height = (Vo^2 × sin^2(angle)) / (2 × acceleration)
Acceleration in this case is due to gravity, which is approximately 9.8 m/s^2.
Height = (13.6687^2 × sin^2(46°)) / (2 × 9.8)
Height ≈ 4.5363 meters
Therefore, the ball goes approximately 4.5 meters above the cliff to the nearest tenth.
PE gained=KE
mgh=1/2 m (19sin46)^2
solve for h, the height gained.