A BALL THROWN WITH A SPEED OF 100M ATTAINS A HEIGHT OF 150 CALCULATE (A) THE TIME OF FLIGHT (B) RANGE (C) THE ANGLE
speed = 100 m/s?
H = 150 m ?
upward trip, t = half flight time
0 = Vi - 9.81 t
t = Vi/9.81
150 = Vi t - (1/2)(9.81) t^2
150 = Vi^2/9.81 - (1/2)(9.81)(Vi^2/9.81^2)
150 = Vi^2 / 4.9
Vi = 27.1 m/s
then t = 27.1/9.81 = 2.76 going up
plus another 2.76 down
---> 5.53 seconds flight time
sin theta = 27.1/100
theta = 15.7 degrees above horizontal
u = 100 cos 15.7
= 96.3 meters/second horizontal
for 5.53 seconds
range = 532 meters down range
To solve this problem, we can use the following equations of projectile motion:
Vertical motion: h = (v₀²sin²θ) / (2g)
Horizontal motion: R = (v₀²sin2θ) / g
Time of flight: T = (2v₀sinθ) / g
where:
h = maximum height attained by the object
R = range or horizontal distance traveled by the object
v₀ = initial velocity
θ = launch angle
g = acceleration due to gravity (approximately 9.8 m/s²)
Let's solve the problem step by step:
(A) Calculate the time of flight:
Using the vertical motion equation, we have:
h = (v₀²sin²θ) / (2g)
Rearranging the equation and substituting the given values:
150m = (100m/s)²sin²θ / (2 * 9.8m/s²)
Rearranging further:
(sin²θ) = (150m * 2 * 9.8m/s²) / (100m/s)²
(sin²θ) = 294 / 10000
Now, take the square root of both sides:
sinθ ≈ √(294 / 10000)
Using a calculator, find the inverse sine (sin⁻¹) of this value to get the angle (θ). Let's assume θ ≈ 16.9°.
Finally, we can calculate the time of flight using the equation:
T = (2v₀sinθ) / g
T = (2 * 100m/s * sin(16.9°)) / 9.8m/s²
Calculating this, we get:
T ≈ 10.21s
Therefore, the time of flight is approximately 10.21 seconds.
(B) Calculate the range:
Using the horizontal motion equation, we have:
R = (v₀²sin2θ) / g
Substituting the given values:
R = (100m/s)² * sin(2 * 16.9°) / 9.8m/s²
Calculating this, we get:
R ≈ 346.86 meters
Therefore, the range is approximately 346.86 meters.
(C) Calculate the angle:
The angle (θ) can be found by rearranging the equation for the time of flight:
T = (2v₀sinθ) / g
Rearranging the equation and substituting the given values, we get:
θ = sin⁻¹((gT) / (2v₀))
Substituting the values in, we have:
θ = sin⁻¹((9.8m/s² * 10.21s) / (2 * 100m/s))
Calculating this, we get:
θ ≈ 16.9°
Therefore, the angle of projection is approximately 16.9 degrees.