Substitute the 2 in for t.
38(2)-16(2)^2
76-64=12
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
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.
A ball thrown vertically upward returns to the starting point in 8 seconds.-- Its velocity was upward for 4 seconds and downward for the other 4 seconds.-- Its velocity was zero at the turning point, exactly 4 seconds after leaving the hand.-- During the first 4 seconds, gravitational acceleration reduced the magnitude of its upward velocity by(9.8 meters/second2) x (4 seconds) = 39.2 meters per second-- So that had to be the magnitude of its initial upward velocity.
velocity is found by dividing the distance with time. In a second the height traveled is found by multiplying the velocity by the time taken and then dividing the answer by two.
Height reached = 3.7 metres.The mass of the ball is not really relevant.
After just over three and a quarter seconds.
Zero
A the highest point its velocity will be zero.
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
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..
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.