A ball is thrown vertically upward with a speed of 1.53 m/s from a point 4.21 m above the ground. Calculate the time in which the ball will reach the ground.
{Distance Above Ground} = D = (4.21) + (1.53)*t + (1/2)*(-9.8)*t^2
{D = 0} ----> {Time To Reach Ground} = 1.096 sec
Is this right?
Listen to my song BOP hit up da likes
What is the equation for this without the numbers slotted in
Yes, your calculation is correct. To find the time it takes for the ball to reach the ground, you can use the kinematic equation for vertical motion:
Distance above ground (D) = initial height (h) + initial vertical velocity (v0) * time (t) + (1/2) * acceleration due to gravity (g) * t^2
In this case, the initial height (h) is 4.21 m, the initial vertical velocity (v0) is 1.53 m/s, and the acceleration due to gravity (g) is -9.8 m/s^2 (negative because it is acting in the opposite direction to the ball's motion).
Setting D to zero (since the ball reaches the ground), you can solve for t:
D = h + v0 * t + (1/2) * g * t^2
0 = 4.21 + 1.53 * t + (1/2) * (-9.8) * t^2
Simplifying the equation, you get a quadratic equation:
4.21 - 1.53 * t - 4.9 * t^2 = 0
Solving this equation, you find that t is approximately 1.096 seconds. So, the ball will reach the ground after approximately 1.096 seconds.
I didn't check your math, but the equation is right. Do this: put the answer in the google search window to calculate it .
(4.21) + (1.53)*1.096 + (1/2)*(-9.8)*(1.096)^2