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If a ball is thrown vertically up with the velocity of 40m s what would be its velocity after two seconds?
Ignoring air resistance and using g = 9.81 ms-2, velocity = 20.38 ms-1.
The height h of a ball thrown into the air with an initial vertical velocity of 48 feet per second from a height of 5 feet above the ground is given by the equation?
At any time 't' seconds after the ball is released,until it hits the ground,h = 5 + 48 t - 16.1 t2
A rock is thrown from a cliff and hits the ground 5 seconds later at a distance of 50 meters from the cliff How high was the cliff?
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.
How can you used trigonometry in projectile motions?
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.,
Is the total time of a ball thrown upwards is it equal to the total time of the downward 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.
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.
What is the velocity of a ball thrown 20 ft in 4 seconds?
In the case of constant velocity (or speed), velocity = distance / time.
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.
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.
What height is reached by a 4.0kg ball that is thrown vertically up into the air with an initial velocity of 8.5ms1 if by the time it is height (h) metres above the starting point it has a velocity of?
Height reached = 3.7 metres.The mass of the ball is not really relevant.
What would happen if the projectile were thrown with a greater velocity?
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.
What is the instantaneous velocity at maximum height when something is thrown straight up?
The instantaneous velocity at the maximum height is zero because the object momentarily stops moving before falling back down due to gravity.
A rock is dropped from the top of a cliff what is the velocity after 2 seconds 5 seconds 10 seconds?
Since the equation for time=sqrt(2h/g) set 2 seconds for time and 9.8 for gravity so 2=sqrt(2h/9.8) clear the sqrt by squaring both sides 4= 2h/9.8 9.8*4 =2h (9.8*4)/2 = height. now that you have the height, you can do v=distance/time v=height from the equation prior/2 seconds i hope that works..
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.
If a ball is thrown vertically up with the velocity of 40m s what would be its velocity after two seconds?
Ignoring air resistance and using g = 9.81 ms-2, velocity = 20.38 ms-1.
The height h of a ball thrown into the air with an initial vertical velocity of 48 feet per second from a height of 5 feet above the ground is given by the equation?
At any time 't' seconds after the ball is released,until it hits the ground,h = 5 + 48 t - 16.1 t2
What is the velocity of a ball thrown upward at 16 feetsec?
The velocity of the ball is 16 feet/sec when it is thrown upward.
