a ball is thrown from a point 100m above the ground with speed 20m/s.find the velocity,direction and magnitude with which it strikes the ground?

To find the velocity, direction, and magnitude with which the ball strikes the ground, we can use the following steps:

Step 1: Determine the time it takes for the ball to reach the ground using the equation of motion:
h = (1/2) * g * t^2

Where:
h = initial height = 100m
g = acceleration due to gravity = 9.8 m/s^2
t = time

Solving for t:
100 = (1/2) * 9.8 * t^2
t^2 = (100 * 2) / 9.8
t^2 = 20.4
t ≈ √20.4
t ≈ 4.5 seconds (rounded to one decimal place)

Step 2: Calculate the final velocity using the equation of motion:
v = u + g * t

Where:
v = final velocity
u = initial velocity = 20 m/s
t = time = 4.5 seconds

v = 20 + 9.8 * 4.5
v ≈ 20 + 44.1
v ≈ 64.1 m/s (rounded to one decimal place)

Step 3: Determine the direction and magnitude of the final velocity.

Since the ball was thrown downward, the direction of the final velocity will also be downward (negative). The magnitude of the final velocity is the absolute value of the velocity:

|v| = |64.1| = 64.1 m/s

Therefore, the velocity with which the ball strikes the ground is approximately -64.1 m/s (downward).

To find the velocity, direction, and magnitude with which the ball strikes the ground, we can use the equations of motion in projectile motion. Let's break down the problem and solve step by step:

Step 1: Define the variables:
- Initial vertical position (y0) = 100m
- Initial vertical velocity (v0y) = 0 m/s (since the ball is thrown horizontally)
- Acceleration due to gravity (g) = 9.8 m/s² (taking downward direction as negative)
- Time taken to reach the ground (t) = ?
- Final vertical velocity (vfy) = ?
- Final horizontal velocity (vfx) =?
- Total velocity (v) = ?

Step 2: Determine the time taken to reach the ground:
Using the equation of motion: y = y0 + v0yt + (1/2)gt², where y = 0 (since it reaches the ground) and y0 = 100m, and v0y = 0 m/s:

0 = 100 + 0*t - (1/2)*9.8*t²
100 = 4.9*t²
t² = 100/4.9
t ≈ √(20.408)
t ≈ 4.515s (rounded to 3 decimal places)

Step 3: Calculate the final vertical velocity:
Using the equation of motion: vfy = v0y + gt, where v0y = 0 m/s and t = 4.515s:

vfy = 0 + 9.8 * 4.515
vfy ≈ 44.325 m/s (rounded to 3 decimal places)

Step 4: Calculate the final horizontal velocity:
Since there is no horizontal acceleration, the final horizontal velocity (vfx) remains the same as the initial horizontal velocity (v0x). Given that the initial speed (v0) is 20 m/s, the final horizontal velocity is:

vfx = 20 m/s

Step 5: Determine the magnitude and direction of the velocity with which the ball strikes the ground:
To find the magnitude of the velocity (v), we can use the Pythagorean theorem:

v = √(vfx² + vfy²)
v = √((20)² + (44.325)²)
v ≈ √(400 + 1966.695)
v ≈ √(2366.695)
v ≈ 48.649 m/s (rounded to 3 decimal places)

To find the direction, we can use trigonometry. The direction of the velocity will be the angle (θ) that the resultant velocity vector makes with the horizontal axis. We can find this angle using the equation:

θ = tan^(-1)(vfy/vfx)
θ = tan^(-1)(44.325/20)
θ ≈ 65.152° (rounded to 3 decimal places)

Therefore, the velocity, direction, and magnitude with which the ball strikes the ground are approximately 48.649 m/s at an angle of 65.152° below the horizontal axis.