A body is projected with a velocity of 200m/s at an angle of 30 above the horizontal. Take g to be 10m/s2 calculate the time taken for the projectile to reach the maximum height.
10 s
Rubbish
Vi = initial speed up = 200 sin 30 = 100 m/s
v = speed up = Vi - g t
at top v = 0
0 = 100 - 10 t
t = 10 seconds to top
by the way the max height Ht results easily
h = 0 + Vi t - (g/2) t^2
so
Htop = 100*10 - 5*100 = 1000-500 = 500 meters
also it takes10 sec upward, another ten down so in air for 20 s
u = horizontal speed = 200 cos 30
so range = 200 cos 30 * 20 = 3464 meters
To calculate the time taken for the projectile to reach the maximum height, you can use the vertical component of the initial velocity and the acceleration due to gravity.
Given:
Initial velocity (u) = 200 m/s
Launch angle (θ) = 30 degrees
Acceleration due to gravity (g) = 10 m/s^2
Step 1: Resolve the initial velocity into vertical and horizontal components.
The vertical component of the initial velocity (u) is given by:
u_vertical = u * sin(θ)
Substituting the given values:
u_vertical = 200 m/s * sin(30°)
u_vertical = 100 m/s
Step 2: Calculate the time taken to reach maximum height.
Using the equation of motion: vf = u + at, where vf is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
At the maximum height, the final vertical velocity (vf) becomes zero, so we have:
0 = u_vertical + (-g) * t
Solving for t:
t = u_vertical / g
Substituting the values:
t = 100 m/s / 10 m/s^2
t = 10 s
Therefore, the time taken for the projectile to reach the maximum height is 10 seconds.