the maximum height attained by the projectile motion is 600m and the range is 500m find
a. angle of projection
b. velocity of projection
c. total time of flight
Vertical problem:
at max height (half the trip)
600 = Vi t - .5*9.8 t^2
.5*9.8 t^2 - Vi t + 600 = 0
also
v = 0
so
0 = Vi -gt
t = Vi/g = Vi/9.8
so
.5*9.8(Vi^2/9.8^2) - Vi^2/9.8 +600 = 0
600 = .5*VI^2/9.8
Vi = 108 m/s initial up speed
t = rising time = 108/9.8 = 11.1 s
total time = 22.2 s
u = 500/22.2 = 22.5 m/s
tan theta = Vi/u = 108/22.5
= 4.8
theta = 78.2 degrees
speed = sqrt(108^2+22.5^2)
check my arithmetic !
To solve this problem, we can use the equations of projectile motion. The variables we will need are:
- θ (angle of projection): This is the angle at which the object is launched.
- v0 (initial velocity): This is the velocity at which the object is launched.
- H (maximum height): This is the maximum height attained by the projectile.
- R (range): This is the horizontal distance covered by the projectile.
- g (acceleration due to gravity): This is a constant, approximately 9.8 m/s².
Now let's solve for the given variables:
a. Angle of Projection (θ):
We can find the angle of projection using the formula:
θ = arctan((g * R) / (2 * v0²))
In this case, R (range) is given as 500m. Let's solve for θ:
θ = arctan((9.8 * 500) / (2 * v0²))
b. Velocity of Projection (v0):
We can find the velocity of projection using the formula:
v0 = sqrt((g * R) / (sin(2θ)))
In this case, R (range) and θ (angle of projection) are given as 500m and the value given from part a, respectively. Let's solve for v0:
v0 = sqrt((9.8 * 500) / (sin(2θ)))
c. Total Time of Flight (t):
The total time of flight can be calculated using the formula:
t = 2 * (v0 * sin(θ)) / g
In this case, v0 (velocity of projection) and θ (angle of projection) are given from parts a and b, respectively. Let's solve for t:
t = 2 * (v0 * sin(θ)) / g
By substituting the values into these formulas, you can find the answers for θ, v0, and t.