A ball thrown with a speed of 100m/s attains a height of 150.calculate
(a)the time of flight
(b)the angle of projection
(c)the range
vertical problem:
goes up 150 m (remarkable throw)
v = Vi - g t
v = 0 at top
Vi = g t = initial up component of speed
h = Vi t - (1/2) g t^2
150 = Vi t - (1/2) g t^2
150 = g t^2 -(1/2) g t^2
300 = g t^2
300/9.81 = t^2
t = 5.53 seconds upward, 11.1 total in air
Vi = 9.81*5.53 = 54.2 meters/sec Upward
= 100 sin angle up from horizon
so angle = sin^-1(54.2/100)
= 32.8 degrees up from level
for range, horizontal problem
speed horizontal = u = 100 cos32.8 forever
u = 84.1 m/s
time in air = 11.1 seconds
so
range = 84.1*11.1 meters
(I think it was an arrow if not a bullet)
Time of flight:
That would be twice the time it takes the ball to fall H = 150 m, which is
t = 2*sqrt(2H/g) = 11.07 s
(You double the fall time to get the total time of flight)
Launch angle:
V*sinA/g = t/2 = 5.533 s
sin A = 0.5422
A = 32.8 degrees
Range:
(Vo^2/g)*sin(2A) = 930 m
To calculate the time of flight, angle of projection, and the range of a ball thrown with a speed of 100 m/s and attains a height of 150 m, we can use the equations of motion for projectile motion.
(a) Time of Flight:
The formula to calculate the time of flight for a projectile is given by:
time = 2 * u * sin(theta) / g
where u is the initial velocity, theta is the angle of projection, and g is the acceleration due to gravity which is approximately 9.8 m/s^2.
Plugging in the given values:
u = 100 m/s
theta = ?
g = 9.8 m/s^2
150 = 2 * 100 * sin(theta) / 9.8
Next, rearrange the equation to solve for sin(theta):
sin(theta) = (150 * 9.8) / (2 * 100)
sin(theta) = 73.5 / 100
theta = arcsin(73.5 / 100)
Using a calculator, we find that theta ≈ 47.6 degrees.
Now we can substitute the angle value back into the equation to find the time:
time = 2 * 100 * sin(47.6) / 9.8
Calculate this to find the time of flight.
(b) Angle of Projection:
We have already calculated the angle of projection in the previous step, which is approximately 47.6 degrees.
(c) Range:
The formula to calculate the range of a projectile is given by:
range = (u^2 * sin(2 * theta)) / g
Plugging in the given values:
u = 100 m/s
theta = 47.6 degrees
g = 9.8 m/s^2
range = (100^2 * sin(2 * 47.6)) / 9.8
Calculate this to find the range.
Therefore, once you solve the equations:
(a) The time of flight is approximately [calculate]
(b) The angle of projection is approximately 47.6 degrees
(c) The range is approximately [calculate]
To calculate the time of flight, angle of projection, and range of a ball thrown with a speed of 100m/s and attains a height of 150m, we can use the equations of projectile motion.
Let's start with the time of flight (T). The time of flight is the total time it takes for the ball to reach its highest point and then return to the same height.
(a) Time of Flight (T):
We know that the initial vertical velocity of the ball is 0 m/s (as it reaches its highest point). So, we can use the equation:
T = 2 * (Vertical displacement) / (Vertical velocity)
Given that the vertical displacement (h) is 150m and the vertical velocity (Vy) is 0m/s, we have:
T = 2 * 150m / 0m/s
However, dividing by 0 is undefined. We need to consider that at the highest point, the vertical velocity is momentarily 0. Hence, the time to reach the highest point is symmetrical to the time taken to descend. So, we can divide the problem into two parts:
Time to reach the highest point = Time to descend = T/2
T = 2 * (Vertical displacement) / (Vertical velocity)
T = 2 * 150m / 100m/s
T = 3 seconds
Therefore, the time of flight is 3 seconds.
(b) Angle of Projection (θ)
To find the angle of projection, we need to use the horizontal (x) and vertical (y) components of the initial velocity.
The horizontal component of velocity (Vx) remains constant throughout the flight. So, we can use the equation:
Vx = Initial velocity * cos(θ)
Given that the initial velocity (V) is 100m/s, we can rearrange the equation to find θ:
θ = arccos(Vx / V)
θ = arccos(100m/s / 100m/s)
θ = arccos(1)
θ = 0 degrees
Therefore, the angle of projection is 0 degrees.
(c) Range (R)
The range of the ball is the horizontal distance covered during the time of flight. We can use the equation:
R = Horizontal velocity * Time of Flight
The horizontal velocity (Vx) can be found using the equation:
Vx = Initial velocity * cos(θ)
Given that the initial velocity (V) is 100m/s and the angle of projection (θ) is 0 degrees, we have:
Vx = 100m/s * cos(0)
Vx = 100m/s * 1
Vx = 100m/s
Now we can calculate the range:
R = 100m/s * 3s
R = 300 meters
Therefore, the range of the ball is 300 meters.