From the given graph, let's find the time it takes the projectile to reach its maximum height.
Given:
θ = 40 degrees
Initial velocity, vo = 250 m/s.
To find the time it takes the projectile to reach maximum height, apply the formula:
[tex]v_y=v_o\sin \theta-gt[/tex]
Where:
vy is the final velocity at maximum height = 0 m/s
vo is the initial velocity
g is acceleration due to gravity = 9.8 m/s²
[tex]0=v_0\sin \theta-gt[/tex]
Rewrite the formula for the time, t:
[tex]t=\frac{v_0\sin \theta}{g}[/tex]
Input values into the formula and solve for t:
[tex]\begin{gathered} t=\frac{250\sin 40}{9.8} \\ \\ t=\frac{160.67}{9.8} \\ \\ t=16.4\text{ s} \end{gathered}[/tex]
Therefore, the time it takes the projectile to reach its maximum height is 16.40 seconds.
ANSWER:
a. 16.40 s
a. 16.40
