What is the speed of a water bomb after 1.75 s if it is released from rest at a height of 110m above the ground and how high off the ground is the ball?
A baseball is hit straight up into the air at an initial velocity of 35 m/s. How long does it take for the
baseball to reach its highest position?
baseball: 35-9.81t = 0
water bomb: v = -9.81t
h = 110 - 4.9t^2
To calculate the speed of the water bomb after 1.75 s, we can use the equation for velocity in free fall:
v = u + at
Where:
v = final velocity
u = initial velocity (in this case, 0 m/s as the water bomb is released from rest)
a = acceleration due to gravity (approximately 9.8 m/s^2)
t = time (1.75 s)
Plugging in the values:
v = 0 + (9.8 * 1.75)
v = 17.15 m/s
Therefore, the speed of the water bomb after 1.75 s is 17.15 m/s.
To calculate how high off the ground the water bomb is after 1.75 s, we can use the equation for displacement in free fall:
s = ut + (1/2)at^2
Where:
s = displacement (vertical distance)
u = initial velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2, negative because the object is moving in the opposite direction of positive displacement)
t = time (1.75 s)
Plugging in the values:
s = 0 * 1.75 + (0.5 * -9.8 * (1.75)^2)
s = 0 + (-8.533)
s = -8.533 m (negative sign indicates it is below the starting height)
Therefore, after 1.75 s, the water bomb is approximately 8.533 m below the starting height.
To determine the speed of a water bomb after 1.75 seconds and how high off the ground the bomb is, we can make use of the equations of motion.
1. First, let's calculate the speed of the water bomb after 1.75 seconds. We can use the kinematic equation:
v = u + at
where:
v = final velocity
u = initial velocity (in this case, 0 m/s as the water bomb is released from rest)
a = acceleration (gravity, approximately -9.8 m/s^2)
t = time (1.75 seconds)
Plugging in the values, we get:
v = 0 + (-9.8 * 1.75)
v = -17.15 m/s
Therefore, the speed of the water bomb after 1.75 seconds is 17.15 m/s, and the negative sign indicates that it is moving downwards.
2. Next, let's calculate how high off the ground the water bomb is. We can use another kinematic equation:
s = ut + (1/2)at^2
where:
s = displacement (height above the ground)
u = initial velocity (0 m/s)
t = time (1.75 seconds)
a = acceleration (gravity, -9.8 m/s^2)
Plugging in the values, we get:
s = 0 * 1.75 + (1/2) * (-9.8) * (1.75)^2
s = 0 + (-8.53)
s = -8.53 m
Therefore, the water bomb is approximately 8.53 meters above the ground after 1.75 seconds, and again the negative sign indicates that it is below the starting height.
Moving on to the baseball question:
To find out how long it takes for the baseball to reach its highest position, we can use the fact that at the highest point of its trajectory, the baseball's final velocity becomes 0 m/s.
1. We can use the formula for finding the final velocity:
v = u + at
where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (35 m/s)
a = acceleration (gravity, approximately -9.8 m/s^2)
To find the time it takes to reach the highest position, we solve the equation with v = 0:
0 = 35 + (-9.8)t
Rearranging the equation, we have:
9.8t = 35
Dividing both sides by 9.8:
t = 35 / 9.8
t ≈ 3.57 seconds
Therefore, it takes approximately 3.57 seconds for the baseball to reach its highest position.