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What is the height of a rock that is thrown upward with an initial speed of 15 meter per second?
Wiki User
∙ 14y ago
We have no idea how big the rock is, and no way to figure it out. But we can calculate that
it reaches 11.48 meters above the ground before it starts falling.
Wiki User
∙ 14y ago
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If an object is thrown upward with an initial velocity of 32 feet per second from 40 feet above the ground write the height equation?
The height, in feet, above the ground at time t, H(t) = 40 + 32*t - 16*t2
A ball is thrown upward at an initial velocity of 70 feet per second from the top of a 345 foot building -- How long before the ball hits the ground below?
By the Way, guys, this is based on the equation H= -16t2+vt+s
What object shoots upward with an initial velocity of 6 meters per second?
anything shot up with that initial velocity. There isn't anything in specific.
What is h equals -5t2 plus 40t?
That's the formula for the height of an object that was tossed upward at a speed of 40 meters per second, after ' t ' seconds . This object has to be something like a canonball, or a baseball pitched by a professional etc. The initial vertical speed of 40 meters per second is almost 90 miles per hour upward !
How fast is a stone moving when it reaches a height of 11 meters if it is vertically thrown upward with a speed of 18 meters per second?
v2 = u2 + 2as where v = current velocity, u = initial velocity, a = acceleration, and s = displacement. Taking a = - 9.8 ms-2 v2 = 182 - (9.8 x 11 x 2) = 108.4 v = 10.4 ms-1
What is the velocity of a ball thrown upward at 16 ft per second?
The initial velocity of the ball thrown upward at 16 ft per second is 16 ft/s.
What relationship exists between the initial velocity and the maximum height reached by an object thrown upward?
Ignoring air resistance, I get this formula:Maximum height of a vertically-launched object = 1.5 square of initial speed/GI could be wrong. In that case, the unused portion of my fee will be cheerfully refunded.
How do you answer this physics question?
A ball is thrown vertically upward with an initial speed of 20m/s. Two second later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24m/s. At what height above the release point will the ball and stone pass each other?
When a ball is throw upward how does the height reached by the ball depend on its initial speed?
The height reached by a ball thrown upward depends on its initial speed: the higher the initial speed, the higher the maximum height reached. This is because a greater initial speed gives the ball more kinetic energy, allowing it to overcome gravity and reach a higher position before gravity brings it back down.
If an object is thrown upward with an initial velocity of 32 feet per second from 40 feet above the ground write the height equation?
The height, in feet, above the ground at time t, H(t) = 40 + 32*t - 16*t2
What is the velocity of a ball thrown upward at 16 feet and seconds?
The initial velocity of the ball is 16 feet per second when thrown upward. The velocity decreases as the ball travels upward due to gravity until it reaches its peak and starts to fall back down.
A ball is thrown horizontally from a window on the second floor of a building What is the vertical component of its initial velocity?
The vertical component of the initial velocity of the ball thrown horizontally from a window is zero. The ball's initial velocity in the vertical direction is influenced only by the force of gravity, not the horizontal throw.
How long will it take the ball to return to earth when it is thrown straight upward at 150' per second at the height of 5'?
OW! Not long enough!
A ball thrown vertically upward returns to the starting point in 8 seconds.find its initial velocity.?
The total time of flight for a ball thrown vertically upwards and returning to its starting point is twice the time taken to reach maximum height. Therefore, the time taken to reach maximum height is 4 seconds. Given that the acceleration due to gravity is -9.8 m/s^2, using the kinematic equation v = u + at, where v is the final velocity (0 m/s at maximum height), u is the initial velocity, a is the acceleration due to gravity, and t is the time, you can solve for the initial velocity. Substituting the values, u = 9.8 * 4 = 39.2 m/s. Therefore, the initial velocity of the ball thrown vertically upward is 39.2 m/s.
The velocity of a body thrown vertically upward is reduced into half in one secondwhat is the maximum height attained by it?
The maximum height attained by the body can be calculated using the formula: height = (initial velocity)^2 / (2 * acceleration due to gravity). Since the velocity is reduced to half in one second, we can calculate the initial velocity using the fact that the acceleration due to gravity is -9.81 m/s^2. Then, we can plug this initial velocity into the formula to find the maximum height reached.
What is tje velocity of a ball thrown upward at 16 ft/sec?
If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t^2. If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t^2.
How high can a ball go?
The maximum height a ball can reach is dependent on factors such as initial velocity, launch angle, air resistance, and gravity. In a vacuum with no air resistance, the ball will continue to rise until gravity slows it down and makes it come back down.
