A softball is thrown from the origin of an x-y coordinate system with an initial speed of 18m/s at an angle of 35 degrees above
1 answer:
Distance = speed × time
After 0.5s , horizontal distance covered is 0.5s × 18m/s = 9m
9m/ y position = cos 35° , y position = 9 ÷ cos 35° = 10.9870m (rounded up to nearest four decimal places)
After t = 1.0s:
Horizontal distance = 18m/s × 1.0s = 18m
y position = = 21.9740m (rounded up to nearest four decimal places)
After t = 1.5s:
Horizontal distance = 18m/s × 1.5s = 27m
y position = = 32.9709m (rounded up to nearest four decimal places)
After time is 2.0s:
Horizontal distance = 18m/s × 2.0s = 36m
y position = = 43.9479m (rounded up to nearest four decimal places)
You might be interested in
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.
