A ball is thrown horizontally from the top of
a building 150 m high. The ball strikes the
ground 62 m horizontally from the point of
release.
What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s
To find the speed of the ball just before it strikes the ground, we can use the equation of motion for horizontal motion:
distance = speed * time
Since the ball is thrown horizontally, the time of flight is the same as if it were dropped from the same height. We can use the equation for vertical motion:
distance = 1/2 * acceleration * time^2
In this case, the initial velocity in the vertical direction is 0, since the ball is only moving horizontally. The acceleration due to gravity is -9.8 m/s^2 because it acts downwards.
Using the equation for distance in vertical motion, we can solve for time:
150 = 1/2 * -9.8 * time^2
Simplifying the equation:
300 = -9.8 * time^2
Dividing by -9.8:
time^2 = -30.61
Since time cannot be negative, we take the positive square root:
time = √30.61 ≈ 5.52 s
Now, we can use this time to find the horizontal speed of the ball. The horizontal distance is given as 62 m. Using the equation for distance in horizontal motion:
62 = speed * time
Speed = 62 / time
= 62 / 5.52
≈ 11.23 m/s
Therefore, the speed of the ball just before it strikes the ground is approximately 11.23 m/s.
To find the speed of the ball just before it strikes the ground, we can use the concept of projectile motion. In this case, since the ball is thrown horizontally, it has no vertical component of velocity initially.
First, let's break down the given information:
- The height of the building (h) = 150 m
- The horizontal distance from the point of release to the point of impact (x) = 62 m
Since the ball is thrown horizontally, the time taken to reach the ground (t) can be found using the formula:
x = (initial horizontal velocity) * time
Since the initial horizontal velocity is constant, we can rearrange the formula to solve for time as follows:
t = x / (initial horizontal velocity)
Since the ball is in free fall, the time taken for it to fall from the top of the building to the ground is the same as the time calculated above. We can use the formula for free fall motion to calculate the time taken for an object to fall from a certain height:
h = (1/2) * g * t^2
Solving the equation for t, we get:
t = sqrt((2 * h) / g)
where g is the acceleration due to gravity, approximately 9.8 m/s^2.
Now that we have the time taken to reach the ground, we can find the initial horizontal velocity.
Using the formula:
x = (initial horizontal velocity) * t
Rearranging the equation, we get:
initial horizontal velocity = x / t
To calculate the speed (magnitude of the velocity) just before the ball strikes the ground, we can use the Pythagorean theorem since only the horizontal velocity is known:
speed = sqrt((initial horizontal velocity)^2)
Plugging in the given values, the calculations are as follows:
t = sqrt((2 * 150 m) / (9.8 m/s^2))
t ≈ 5.06 s
initial horizontal velocity = 62 m / 5.06 s ≈ 12.24 m/s
speed = sqrt((12.24 m/s)^2)
speed ≈ 12.24 m/s
Therefore, the speed of the ball just before it strikes the ground is approximately 12.24 m/s.
To find the speed of the ball just before it strikes the ground, we can use the equations of motion.
The ball was thrown horizontally from the top of the building and there is no vertical acceleration. Therefore, the time it takes for the ball to hit the ground is the same as the time it would take for an object to fall vertically from a height of 150 m.
Using the equation for vertical displacement, s = ut + (1/2)gt^2, where u is the initial vertical velocity (zero in this case), g is the acceleration due to gravity (-9.8 m/s^2), and s is the vertical displacement (150 m), we can rearrange the equation to solve for time, t.
150 = (1/2)(-9.8)t^2
t^2 = -150 / (-4.9)
t^2 = 30.61
t ≈ √30.61
t ≈ 5.53 seconds
Therefore, it takes approximately 5.53 seconds for the ball to hit the ground.
Now, to find the horizontal velocity of the ball, we use the formula v = d/t, where v is the velocity, d is the horizontal distance (62 m), and t is the time (5.53 s).
v = 62 / 5.53
v ≈ 11.21 m/s
So, the speed of the ball just before it strikes the ground is approximately 11.21 m/s.