Answer:
The maximum height the pebble reaches is approximately;
A. 6.4 m
Explanation:
The question is with regards to projectile motion of an object
The given parameters are;
The initial velocity of the pebble, u = 19 m/s
The angle the projectile path of the pebble makes with the horizontal, θ = 36°
The maximum height of a projectile, [tex]h_{max}[/tex], is given by the following equation;
[tex]h_{max} = \dfrac{\left (u \times sin(\theta) \right)^2}{2 \cdot g}[/tex]
Therefore, substituting the known values for the pebble, we have;
[tex]h_{max} = \dfrac{\left (19 \times sin(36 ^{\circ}) \right)^2}{2 \times 9.8} = 6.3633894140470403035477570509439[/tex]
Therefore, the maximum height of the pebble projectile, [tex]h_{max}[/tex] ≈ 6.4 m.
