a man stand on the top of the building 490 m above the ground.he throws a stone horizontally with speed15 m/s.find the time taken by the stone to reach the ground &the speed with which it hits the ground
h=gt²/2
t=sqrt(2h/g)
v(y) = gt
v(x) =15 m/s
v= sqrt{v(x)²+v(y)²}
A ball is thrown vertically upward with an
initial speed of 28 m/s. Then, 1.7 s later, a
stone is thrown straight up (from the same
initial height as the ball) with an initial speed
of 33.3 m/s.
How far above the release point will the ball
and stone pass each other? The acceleration
of gravity is 9.8 m/s^2
Answer in units of m
To find the time taken by the stone to reach the ground and the speed with which it hits the ground, we can use the principles of projectile motion.
1. First, let's analyze the horizontal motion of the stone. Since the stone is thrown horizontally, there is no initial vertical velocity. Hence, the horizontal motion has constant velocity.
- The horizontal velocity (v_x) of the stone remains constant at 15 m/s throughout its motion.
Now, let's focus on the vertical motion of the stone:
2. We know that the stone is released from a height of 490 m above the ground, and there is no initial vertical velocity. The only force acting on the stone is gravity.
3. The vertical motion can be analyzed using the equation: h = ut + (1/2)gt^2, where
- h is the height of the stone (490 m),
- u is the initial vertical velocity (0 m/s),
- g is the acceleration due to gravity (approximately 9.8 m/s²), and
- t is the time taken by the stone to reach the ground (which we need to find).
4. In this case, the equation can be simplified to: 490 = 0 + (1/2)(9.8)t^2.
5. Rearrange the equation to solve for t:
- Divide both sides of the equation by (1/2)(9.8):
490 / [(1/2)(9.8)] = t^2.
- Simplify the equation further:
490 / 4.9 = t^2.
100 = t^2.
6. Take the square root of both sides to find the time (t):
- √(100) = √(t^2).
- t = ±10.
Since time cannot be negative in this context, the time taken by the stone to reach the ground is t = 10 seconds.
Now, let's find the speed with which the stone hits the ground:
7. Since the stone is thrown horizontally, its horizontal velocity remains constant at 15 m/s.
8. The vertical velocity (v_y) of the stone when it hits the ground can be found using the equation: v_y = u + gt, where
- v_y is the final vertical velocity,
- u is the initial vertical velocity (0 m/s),
- g is the acceleration due to gravity (approximately 9.8 m/s²), and
- t is the time taken by the stone to reach the ground (10 seconds).
9. Plug in the values: v_y = 0 + (9.8)(10).
- v_y = 98 m/s.
Therefore, the speed with which the stone hits the ground is 98 m/s.