Answer:
19.4 m (3 s.f.)
Explanation:
When a body is projected through the air with initial speed u, at an angle of θ to the horizontal, it will move along a curved path.
Therefore, trigonometry can be used to resolve the body's initial velocity into its vertical and horizontal components:
- Horizontal component of u (x) = u cos θ
- Vertical component of u (y) = u sin θ
Because the projectile is modeled as moving only under the influence of gravity, the only acceleration the projectile will experience will be acceleration due to gravity (a = 9.8 m/s²).
[tex]\boxed{\begin{minipage}{9 cm}\underline{The Constant Acceleration Equations (SUVAT)}\\\\s = displacement in m (meters)\\u = initial velocity in m s$^{-1}$ (meters per second)\\v = final velocity in m s$^{-1}$ (meters per second)\\a = acceleration in m s$^{-2}$ (meters per second per second)\\t = time in s (seconds)\\\\When using SUVAT, assume the object is modeled\\ as a particle and that acceleration is constant.\end{minipage}}[/tex]
If a javelin is thrown at 28.5 m/s from flat ground at an angle of 43.2° then:
- Horizontal component of u = 28.5 cos 43.2°
- Vertical component of u = 28.5 sin 43.2°
When the javelin reaches its maximum height, the vertical component of its velocity will momentarily be zero.
Resolving vertically, taking up as positive:
[tex]s = s_y,\quad u=28.5 \sin (43.2^{\circ}), \quad v = 0, \quad a=-9.8[/tex]
[tex]\begin{aligned}\textsf{Using }\:v^2 & = u^2+2as:\\\\\implies 0^2 & = [28.5 \sin (43.2^{\circ})]^2+2(-9.8)s\\0 & = 380.624...-19.6s\\ 19.6s & = 380.624...\\ s & = 19.41960...\\ s & = 19.4\:\:\sf m\:(3\:s.f.)\end{aligned}[/tex]
Therefore, the maximum height of the javelin is 19.4 m (3 s.f.).
Learn more about constant acceleration equations here:
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