When an object is thrown at an angle from the horizontal, the path followed by the body is called the projectile motion. The duration in seconds that the object is above the height of 14 m is 4.698 s.
Speed is the time rate at which velocity is changing.
Vertical speed component is
V₀y = 52sin 30°= 26 m/s
Given is the height h=14m
Using second equation of motion,
S=ut+ 1/2 at²
Substituting the values, we get
14 = 26t - 4.9t²
4.9t² - 26t +14 =0
Solving the quadratic equation, we get the time as
taking the positive sign, t =4.698 s
taking the negative sign , t = 0.6082 s
Thus, the duration in seconds that the object is above the height of 14 m is 4.698s.
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