7. A projectile is fired with an initial

To solve the given problem, we can use the equations of projectile motion. Let's break it down into four parts:
(a) Maximum height reached by the projectile:
The maximum height reached by the projectile can be determined using the formula for vertical displacement:
$h_{\text{max}}=\frac{v_0^2{\sin }^2(\theta )}{2g}$
where:
- $h_{\text{max}}$ is the maximum height
- $v_0$ is the initial speed (65 m/s)
- $\theta$ is the launch angle (35 degrees)
- $g$ is the acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the given values into the equation, we have:
$h_{\text{max}}=\frac{{65}^2{\sin }^2(35)}{2\times 9.8}$
Calculating this expression will give us the maximum height reached by the projectile.
(b) Total time in the air:
The total time in the air can be determined using the formula for the total flight time:
$t_{\text{total}}=\frac{2v_0\sin (\theta )}{g}$
where:
- $t_{\text{total}}$ is the total time in the air
- $v_0$ is the initial speed (65 m/s)
- $\theta$ is the launch angle (35 degrees)
- $g$ is the acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the given values into the equation, we have:
$t_{\text{total}}=\frac{2\times 65\sin (35)}{9.8}$
Calculating this expression will give us the total time the projectile remains in the air.
(c) Total horizontal distance covered (range):
The total horizontal distance covered, also known as the range, can be determined using the formula:
$R=\frac{v_0^2\sin (2\theta )}{g}$
where:
- $R$ is the total horizontal distance covered (range)
- $v_0$ is the initial speed (65 m/s)
- $\theta$ is the launch angle (35 degrees)
- $g$ is the acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the given values into the equation, we have:
$R=\frac{{65}^2\sin (2\times 35)}{9.8}$
Calculating this expression will give us the total horizontal distance covered by the projectile.
(d) Velocity of the projectile 1.50 s after firing:
The velocity of the projectile at any given time can be determined using the following equations:
$v_x=v_0\cos (\theta )$ (horizontal component)
$v_y=v_0\sin (\theta )-gt$ (vertical component)
where:
- $v_x$ is the horizontal velocity
- $v_y$ is the vertical velocity
- $v_0$ is the initial speed (65 m/s)
- $\theta$ is the launch angle (35 degrees)
- $g$ is the acceleration due to gravity (approximately 9.8 m/s^2)
- $t$ is the time after firing (1.50 s)
Using the given values, we can calculate the horizontal
and vertical components of the velocity. The magnitude of the velocity can be determined using the Pythagorean theorem:
$v=\sqrt{v_x^2+v_y^2}$
Calculating these expressions will give us the velocity of the projectile 1.50 s after firing.