Complete all four parts. 15 points. Will give brainliest! Show work!

A. 5.08 secs.

B. 10.16 secs.

C. 126.50 m.

D. 373.36 m

Explanation:

Data obtained from the question include the following:

Initial velocity (u) = 65 m/s

Angle of projection θ = 50°

A. Determination of the time taken to reach the peak.

Initial velocity (u) = 65 m/s

Angle of projection θ = 50°

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =.?

t = u•Sine θ/g

t = (65 × Sine 50) /9.8

t = 5.08 secs.

B. Determination of the total time spent by the ball in air.

Time (t) taken to reach the peak = 5.08 secs.

Total time (T) spent by the ball in air =?

T = 2t

T = 2 × 5.08

T = 10.16 secs

Therefore, the total time spent by the ball in air is 10.16 secs.

C. Determination of the maximum height.

Initial velocity (u) = 65 m/s

Angle of projection θ = 50°

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (H) =..?

H = u²•Sine² θ / 2g

H = 65² × (Sine 50)² / 2 × 9.8

H = 4225 × (Sine 50)² /19.6

H = 126.50 m

Therefore, the maximum height reached by the ball is 126.50 m.

D. Determination of the horizontal distance travelled by the ball.

Initial velocity (u) = 65 m/s

Angle of projection θ = 50°

Acceleration due to gravity (g) = 9.8 m/s²

Horizontal distance (R) =..?

R = u²•Sine 2θ / g

R = 65² × Sine (2×30) / 9.8

R = (4225 × Sine 60) / 9.8

R = 373.36 m

Therefore, the horizontal distance travelled by the ball is 373.36 m

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.