Question

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 the horizontal part b find the y positions of the softball at times t=0.50 s t=1.0 s t=1.5 s, t=2.0 s

Answer 1

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 = $\frac{18m}{cos 35°}$ = 21.9740m (rounded up to nearest four decimal places)

After t = 1.5s:

Horizontal distance = 18m/s × 1.5s = 27m

y position = $\frac{27}{cos 35°}$ = 32.9709m (rounded up to nearest four decimal places)

After time is 2.0s:

Horizontal distance = 18m/s × 2.0s = 36m

y position = $\frac{36}{cos 35°}$ = 43.9479m (rounded up to nearest four decimal places)