A ball is thrown at 21m/s at 30degree above the horizontal from the top of a roof 16m high.calculate the time of flight

The initial vertical velocity component is

Vyo = 21 sin 30 = 10.5 m/s.

Height above ground (y) is given by the equation:
y = y(0) + Vyo*t - (g/2)t^2
= 15 + 10.5 t - 4.9 t^2
Set y = 0 and solve for t. Take the positive one of the two roots of the quadratic equation.
4.9 t^2 -10.5 t -10 = 0

t = (1/9.8)*[10.5 + sqrt(110.25 +196)]
= 2.86 seconds

find range

Why did the ball throw a party on the roof? It wanted to have a "high" time! But in all seriousness, let's calculate the time of flight for the ball.

We can break down the initial velocity of the ball into horizontal and vertical components. The vertical component is given by V_y = V₀ * sin(θ), where V₀ is the initial velocity (21 m/s) and θ is the launch angle (30 degrees).

Using the equation for vertical displacement, we can find the time it takes for the ball to reach its maximum height (which is when the ball's vertical velocity becomes zero). The equation is Δy = V_y₀ * t + (1/2) * g * t², where Δy is the vertical displacement, V_y₀ is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (approximated as 9.8 m/s²).

In this case, the initial vertical velocity is V_y = 21 m/s * sin(30 degrees), and the vertical displacement is Δy = 16 m (the height of the roof). Solving for t gives us:

16 m = (21 m/s * sin(30 degrees)) * t - (1/2) * 9.8 m/s² * t²

Simplifying this equation will give us the time it takes for the ball to reach its maximum height. But rather than calculating it step-by-step here, why don't you give it a try? I'll be here to help if you need any assistance!

To calculate the time of flight of the ball, we can use the following equation:

time of flight = (2 * initial velocity * sine of angle) / acceleration due to gravity

Given:
Initial velocity (u): 21 m/s
Angle (θ): 30 degrees
Height (h): 16 m

Acceleration due to gravity (g) is approximately 9.8 m/s².

First, let's convert the angle from degrees to radians:

θ_radians = θ * (π/180)
θ_radians = 30 * (π/180)
θ_radians ≈ 0.5236 radians

Next, we can calculate the time of flight using the equation mentioned above:

time of flight = (2 * u * sin(θ_radians)) / g
time of flight = (2 * 21 * sin(0.5236)) / 9.8
time of flight ≈ 2.39 seconds

Therefore, the time of flight of the ball is approximately 2.39 seconds.

To calculate the time of flight of the ball, we can use the kinematic equations of motion.

The given information is as follows:
Initial velocity (u) = 21 m/s
Initial height (h) = 16 m
Angle of projection (θ) = 30 degrees

The horizontal and vertical components of the initial velocity can be calculated as follows:
Horizontal component: ux = u * cos(θ)
Vertical component: uy = u * sin(θ)

Since the ball is projected horizontally, there is no initial vertical velocity (uy = 0). Therefore, we can use the equation for vertical displacement to calculate the time of flight.

The equation for vertical displacement from a given initial height (h) and initial vertical velocity (uy = 0) is:
h = ut + (1/2) * g * t^2

Rearranging the equation to solve for time (t):
h = (1/2) * g * t^2
2h = g * t^2
t^2 = (2h) / g
t = sqrt((2h) / g)

Where g is the acceleration due to gravity, approximately 9.8 m/s².

Plugging in the given values:
h = 16 m
g = 9.8 m/s²

t = sqrt((2 * 16) / 9.8)
t = sqrt(32 / 9.8)
t ≈ 2.04 seconds

Therefore, the time of flight of the ball is approximately 2.04 seconds.