Eclipse87
Substitute the 2 in for t.
38(2)-16(2)^2
76-64=12
Wiki User
∙ 14y agoIgnoring air resistance and using g = 9.81 ms-2, velocity = 20.38 ms-1.
At any time 't' seconds after the ball is released,until it hits the ground,h = 5 + 48 t - 16.1 t2
Assuming you throw the rock horizontally off the cliff it drops down at the acceletrtion of gravity. height= 1/2 gt^2 With g = 9.8 m/sec and t = 5 seconds we have height = (1/2) (9.8)(5)(5) = 122.5 meters notice it has nothing todo with the 50 meter distance, which depends on the horizontal velocity.
There's no such thing as "time of the downward velocity", but I think I get the sense of your question. If the effects of air resistance can be disregarded, then any object thrown upwards spends half of its time rising, and the identical amount of time falling back to the height of your hand when you let it go.
when a body is thrown at an angle in a projectile motion, the vertical component of the velocity is vcos(B) ..where v is the velocity at which the body is thrown and B represents the angle at which it is thrown.Similarly horizontal component is vsin(B). these components are useful in determining the range of the projectile ,the maximum height reached,time of ascent,time of descent etc.,
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.
In the case of constant velocity (or speed), velocity = distance / time.
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.
Height reached = 3.7 metres.The mass of the ball is not really relevant.
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.
If the projectile is thrown with a greater velocity, it would travel further and potentially reach a higher peak height. The increased velocity would also result in a shorter flight time and the projectile hitting the ground with a greater impact force.
The instantaneous velocity at the maximum height is zero because the object momentarily stops moving before falling back down due to gravity.
The velocity after 2 seconds, 5 seconds, and 10 seconds can be calculated using the formula v = gt, where g is the acceleration due to gravity (approximately 9.81 m/s^2). After 2 seconds, the velocity is 19.62 m/s downward. After 5 seconds, the velocity is 49.05 m/s downward. After 10 seconds, the velocity is 98.1 m/s downward.
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.
Ignoring air resistance and using g = 9.81 ms-2, velocity = 20.38 ms-1.
At any time 't' seconds after the ball is released,until it hits the ground,h = 5 + 48 t - 16.1 t2
If you ignore air resistance, then they will reach their maximum height at the same time. In order not to ignore air resistance, you would need to know their shapes.