An object is thrown upward with a speed of 21.5 m/s. How high above the projection point is it after 2.06 s?
h = vₒ•t - g•t²/2
Thank you. This was really helpful. I was never given this equation.
The only equation I got was:
y= y_o + v_yo - 1/2(gt^2)
James. It is exactly the same equation.
1/2 (a)= a/2
Well, then I made a calculation error because the two equation gave me a different answer.
To solve this problem, we can use the equations of motion for an object in free fall. The key equation in this case is the equation for the height of an object as a function of time:
h(t) = h0 + v0t - (1/2)gt^2
where:
h(t) is the height of the object at time t.
h0 is the initial height of the object (which is zero in this case since it is thrown upward from the ground).
v0 is the initial velocity of the object (21.5 m/s).
g is the acceleration due to gravity (approximately 9.8 m/s^2).
Given that the initial height (h0) is 0, the initial velocity (v0) is 21.5 m/s, and the time (t) is 2.06 s, we can now substitute these values into the equation:
h(2.06) = 0 + (21.5)(2.06) - (1/2)(9.8)(2.06)^2
h(2.06) = 0 + 44.29 - 10.06
h(2.06) = 34.23 meters
Therefore, the object is approximately 34.23 meters above the projection point after 2.06 seconds.