Answer:
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
