A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of 3m/s² for 0.5 minute. If the maximum height reached by it is 80m then the angle of projection is[g=10m/s²]
v=u+at
v=0+3×30 (0.5×60=30)
v=90
u cosx=90......... eq (1)
H=u^2sin^2x÷2g=80 (given)
u sinx=40 (solving)......eq (2)
eq (1)÷eq (2)
tanx=4/9
x=tan^-1 (4/9)
To find the angle of projection, we need to analyze the motion of the projectile using the given information. We will break down the problem into different components and solve them step by step.
Step 1: Determine the initial velocity components:
The given information tells us that the horizontal component of the velocity of the projectile is equal to the velocity it would acquire if it had moved from rest with uniform acceleration for 0.5 minutes. Let's calculate this velocity:
time = 0.5 minutes = 0.5 * 60 seconds = 30 seconds
acceleration = 3 m/s²
Using the formula: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can calculate the horizontal velocity:
u_horizontal = u = 0 + (3 m/s² * 30 s) = 90 m/s
Step 2: Determine the vertical velocity component:
To find the vertical component of the initial velocity, we need to consider the maximum height reached by the projectile. At the maximum height, the vertical velocity component will be zero.
Using the formula: v = u + gt, where g is the acceleration due to gravity (10 m/s²), and v is the final velocity (0 at the maximum height), we can calculate the vertical component of the initial velocity:
0 = u_vertical + (10 m/s² * t)
u_vertical = -10 m/s * t
Since the time, t, is the same for both horizontal and vertical velocities, we can express u_vertical in terms of u_horizontal:
u_vertical = -10 m/s * t = -10 m/s * 30 s = -300 m/s
Step 3: Calculate the resultant velocity:
The initial velocity components, u_horizontal and u_vertical, form a right-angled triangle. The resultant velocity (u) can be obtained using the Pythagorean theorem:
u = √(u_horizontal² + u_vertical²)
u = √(90 m/s)² + (-300 m/s)²)
u = √(8100 m²/s² + 90000 m²/s²)
u = √(98100 m²/s²)
u ≈ 313.17 m/s
Step 4: Calculate the angle of projection:
To find the angle of projection (θ), we will use trigonometric ratios. In particular, the tangent of the angle is equal to the vertical component of velocity divided by the horizontal component:
tan(θ) = u_vertical / u_horizontal
θ = tan⁻¹(u_vertical / u_horizontal)
θ = tan⁻¹(-300 m/s / 90 m/s)
θ = tan⁻¹(-3.33)
Using a scientific calculator, θ ≈ -72.3° (rounded to one decimal place)
Note: The negative sign indicates that the angle is measured below the horizontal axis (downwards).
Therefore, the angle of projection is approximately -72.3°.
Vh = 3 m/s² * 30 s = 90 m/s
½ m Vv² = m g h
... Vv = √(2 g h) = √( 2 * 10 * 80)
tan(Θ) = Vv / Vh