Question

The launching speed of certain projectile is 5 times it has at its maximum height .Its angle of projection is

Answer 1

Answer on Question # 71482, Physics / Mechanics | Relativity

Question

The launching speed of certain projectile is 5 times it has at its maximum height. Its angle of projection is

Solution

Let initial speed is $v_{0}$ . Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so the magnitudes of the components of the speed $v_{0}$ are

$$v _ {x 0} = v _ {0} \cos \theta , v _ {y 0} = v _ {0} \sin \theta$$

where $\theta$ is the direction of velocity, as shown in Figure. Initial values are denoted with a subscript 0.

Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion have the following forms:

Horizontal Motion $(a_{x} = 0)$ $v_{x} = v_{0x}$ velocity is a constant

Vertical Motion (assuming positive is up $a_{y} = -g = -9.8m / s^{2}$ ) $v_{y} = v_{y0} - gt$

The highest point in any trajectory is reached when $v_{y} = 0$ , so at the maximum height the speed $v$ of the projectile is $v = v_{x0}$ . By the condition of problem $v_{0} = 5v$ , or $v_{0} = 5v_{x0}$ . Since

$$\cos \theta = \frac {v _ {x 0}}{v _ {0}}$$

we get

$$\cos \theta = \frac {v _ {x 0}}{v _ {0}} = \frac {v _ {x 0}}{5 v _ {x 0}} = \frac {1}{5}$$

so

$$\theta = \cos^ {- 1} (0. 2) = 7 8. 5 {}^ {\circ}$$

Answer: $\theta = 78.5{}^{\circ}$

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