A projectile is fired from the ground level with initial velocity of 90m/s at an angle of 36degre from horizontal. find the magnitude and direct of projectile 5 secs after firring
u = 90 cos 36
v = 90 sin 36 - 9.8 t
magnitude = sqrt (u^2+v^2)
tan angle above horizontal = v/u
Whats the earth's gravitational field strenghth
To find the magnitude and direction of the projectile 5 seconds after firing, we need to break down the problem into its horizontal and vertical components.
First, let's find the horizontal component. We can use the initial velocity and angle of projection to find the horizontal velocity.
Horizontal component of velocity (Vx) = Initial velocity (V) * cosine(angle)
Vx = 90 m/s * cos(36°) = 90 m/s * 0.809 = 72.81 m/s
Now, let's find the vertical component. We can use the initial velocity and angle of projection to find the initial vertical velocity, and then use the equations of motion to find the vertical position of the projectile after 5 seconds.
Vertical component of velocity (Vy) = Initial velocity (V) * sine(angle)
Vy = 90 m/s * sin(36°) = 90 m/s * 0.5878 = 52.91 m/s
Using the equation:
Vertical position (h) = Initial vertical velocity (Vy) * time (t) - 0.5 * acceleration due to gravity (g) * time squared (t^2)
h = Vy * t - 0.5 * g * t^2
where g = 9.8 m/s^2 (acceleration due to gravity)
h = 52.91 m/s * 5 s - 0.5 * 9.8 m/s^2 * (5 s)^2
h = 264.55 m - 0.5 * 9.8 m/s^2 * 25 s^2
h = 264.55 m - 0.5 * 9.8 m/s^2 * 625 s^2
h = 264.55 m - 0.5 * 9.8 m/s^2 * 625 s^2
h = 264.55 m - 30.49 m
h ≈ 234.06 m
The magnitude of the projectile's position after 5 seconds is approximately 234.06 meters.
To find the direction of the projectile, we need to find the angle it makes with the horizontal. We can use the inverse tangent function (tan^-1) to find the angle.
Direction = tan^-1(Vy/Vx)
Direction = tan^-1(52.91 m/s / 72.81 m/s) ≈ 38.3°
Therefore, the magnitude of the projectile 5 seconds after firing is approximately 234.06 meters, and the direction is approximately 38.3° above the horizontal.