A body is projected upwards at an angle of 30¤ with the horiztal of an initial speed of 200ms-1 a)in how many second will it reach the ground b)how far from the point of projection will it stirke?
I do not understand the solving
Please Help Me Out.
there is an equation for range (distance to point of impact)
where v is the launch velocity and Θ is the launch angle
r = v^2 sin(2Θ) / g
the time of flight is ... 2 v sin(Θ) / g
To solve this problem, we can use the principles of projectile motion and break it down into two components: horizontal and vertical.
a) In how many seconds will it reach the ground?
First, we need to find the time it takes for the body to reach the highest point of its trajectory. Since the body is projected upwards, the vertical component of its initial velocity is given by:
Initial vertical velocity (Vy) = Initial speed (200 m/s) * sin(angle)
V_y = 200 m/s * sin(30°)
V_y = 100 m/s
Using the equation for vertical motion:
Vertical displacement (Sy) = (V_y^2 - V_0y^2) / (2 * g)
where V_0y is the initial vertical velocity, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the body is projected upwards, the vertical displacement at the highest point of its trajectory is zero. So, we can calculate the time of flight (t) to reach the highest point:
0 = (100^2 - 200^2 * sin^2(30°)) / (2 * 9.8)
0 = (10000 - 200^2 * (1/4)) / (2 * 9.8)
0 = (10000 - 10000/4) / (2 * 9.8)
0 = (10000 - 2500) / (2 * 9.8)
0 = 7500 / (2 * 9.8)
0 = 383.673 seconds
The time taken to reach the highest point is 383.673 seconds. However, the body will take the same amount of time to come down from the highest point to the ground. Therefore, the total time of flight is twice that:
Total time of flight = 2 * 383.673
Total time of flight = 767.346 seconds
So, it will take approximately 767.346 seconds for the body to reach the ground.
b) How far from the point of projection will it strike?
To find the horizontal distance covered by the body, we can use the equation:
Horizontal distance (Sx) = Initial horizontal velocity (Vx) * Time of flight (t)
The initial horizontal velocity (Vx) can be calculated using:
Initial horizontal velocity (Vx) = Initial speed (200 m/s) * cos(angle)
V_x = 200 m/s * cos(30°)
V_x = 173.205 m/s
Using the time of flight we calculated earlier, we can find the horizontal distance covered:
S_x = 173.205 m/s * 767.346 seconds
S_x = 132,858.605 meters
So, the body will strike approximately 132,858.605 meters (or 132.86 km) from the point of projection.