Answer:
1.53 seconds (2 d.p.)
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).
Constant Acceleration Equations (SUVAT)
[tex]\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}[/tex]
When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.
If an object is thrown at 30 m/s from flat ground at an angle of 30° then:
- Horizontal component of u = 30 cos 30°
- Vertical component of u = 30 sin 30°
When the object reaches its maximum height, the vertical component of its velocity will momentarily be zero.
Resolving vertically, taking up as positive:
- [tex]u = 30 \sin 30^{\circ}, \quad v = 0, \quad a = -9.8[/tex]
[tex]\begin{aligned}\textsf{Using}\;\; v &=u+at:\\\\\implies 0 &=30 \sin 30^{\circ}+(-9.8)t\\0 &=15-9.8t\\9.8t &=15\\t &=\dfrac{15}{9.8}\\t&=1.53\;\; \sf s\;(2 \; d.p.)\end{aligned}[/tex]
Therefore, the length of time the object takes to reach the highest point of its flight is 1.53 seconds (2 d.p.).
