A partical is projected vertically upward with speed 20 m/s. What will be the time after particle is at a height 15m above the ground( g=15 m/s*s)
h=ho+vi*t-1/2 g t^2
15=0+20t-4.9t^2
Put it in the form of a quadratic equation
at^2+bt+c=0 and use the quadratic equation.
If velocity v of particle moving in straight line is related with distance travelled S as v=2(1+S)^1/2. Find the acceleration of the particle.
The actual value of g on Earth is 9.8 m/s^2, which is what BobPursley assumed.
I have no idea you were asked to assume g = 15 m/s^2.
Your follow-up question should have been posted separately.
Use a = dV/dt = dV/ds * ds/dt = V* dV/ds
To find the time it takes for the particle to reach a height of 15m above the ground, we can use the kinematic equation for vertical motion:
h = u*t + (1/2)*g*t^2
Where:
h = height of the particle above the ground (15m)
u = initial upward velocity of the particle (20 m/s)
g = acceleration due to gravity (15 m/s^2)
t = time
Rearranging the equation to solve for t:
(1/2)*g*t^2 + u*t - h = 0
We now have a quadratic equation in terms of t. We can solve this equation using the quadratic formula:
t = (-u ± √(u^2 - 4*(1/2)*g*(-h))) / (2*(1/2)*g)
Simplifying further:
t = (-u ± √(u^2 + 2*g*h)) / g
Substituting the given values:
t = (-20 ± √(20^2 + 2*15*15)) / 15
Calculating inside the square root:
t = (-20 ± √(400 + 450)) / 15
t = (-20 ± √850) / 15
Using a calculator to find the square root:
t ≈ (-20 ± 29.14) / 15
This gives us two possible solutions for t. To determine which one is valid, we use the one that gives a positive value for t since time cannot be negative in this context.
Using the positive value:
t ≈ (-20 + 29.14) / 15 ≈ 0.61 seconds
Therefore, it will take approximately 0.61 seconds for the particle to reach a height of 15m above the ground.