Question

for which angle the vertical height of projectile is the maximum?
a 0 degree
b 45 degree
c 60 degree
d 90 degree

Explanation:

Step1: Recall projectile - height formula

The formula for the maximum height \(H\) of a projectile launched with initial velocity \(v_0\) at an angle \(\theta\) is \(H=\frac{v_0^{2}\sin^{2}\theta}{2g}\), where \(g\) is the acceleration due to gravity.

Step2: Analyze the function of \(\theta\)

We know that the range of the sine - function is \([- 1,1]\), and we want to maximize \(\sin^{2}\theta\). The maximum value of \(\sin^{2}\theta\) occurs when \(\sin\theta = 1\).

Step3: Find the value of \(\theta\)

Since \(\sin\theta = 1\) when \(\theta = 90^{\circ}\), the maximum height of the projectile is achieved when the angle of projection \(\theta = 90^{\circ}\).

Answer:

d. 90 degree