A ball is thrown vertically upward from a height of 2 m with initial velocity of 20 m/s (how high with the ball go? assume the acceleration is a(t)= -9.8 m/s)
h(t) = -4.9t^2 + 20t + 2
as with all quadratics, the vertex is at t = -b/2a = 20/9.8
so find h there.
Pls helppppp
But where did you get the 4.9??
it's 1.2 g, which they gave you
h(t) = 1/2 gt^2 + v0*t + h0
better review the basic formulas of ballistic trajectories.
In this question we start with the equation for the acceleration of the ball, which is -32 feet per second. Then, we take the antiderivative of this function in order to find our equation for velocity, which is -32t + C. In order to find out what C is, we use our initial velocity, which is 40. Thus, our equation for velocity is -32t + 40. Then, we need to figure out what the time is when the height of the ball is 0. Thus, we must take the antiderivative of the velocity equation, which results in -16t^2 + 40t + C. Again, in order to figure out what C is, we simply use the initial height of 500. This gives us a positional equation of -16t^2 + 40t + 500. Then, in order to figure out at what time the ball hits the ground, we set the positional equation to 0. This gives us the equation 0 = 16t^2 + 40t + 500. Then, once we factor this using the quadratic formula, we get (5/4) +/- (5*sqrt(21))/4, or t = ā4.47822 and t = 6.97822. Now, as time cannot be negative, then t must equal 6.97822. Then we simply put this value into our velocity equation, and get -183.303 as our final answer. (rounded)
To find out how high the ball will go, we can use the equations of motion for an object thrown vertically with constant acceleration. Specifically, we can use the equation:
h(t) = h0 + v0 * t + (1/2) * a * t^2,
where:
- h(t) is the height of the ball at time t,
- h0 is the initial height (2 m in this case),
- v0 is the initial velocity (20 m/s in this case),
- a is the acceleration due to gravity (-9.8 m/s^2 in this case), and
- t is the time.
Since the ball is thrown vertically upward, the initial velocity is positive, and the acceleration due to gravity is negative (as it opposes the motion). Therefore, we need to consider the upward motion first until the ball reaches its highest point, and then the downward motion until it reaches the initial height.
To find the time it takes for the ball to reach its highest point, we can use the equation for vertical motion:
v = v0 + a * t.
At the highest point, the velocity is 0 since the ball momentarily stops. So, we can solve the equation for t:
0 = 20 - 9.8 * t.
Simplifying the equation, we have:
9.8 * t = 20.
Dividing both sides by 9.8, we find:
t ā 2.041 s.
Now, we can substitute this value of t into the equation for height:
h(t) = 2 + 20 * 2.041 - 0.5 * 9.8 * 2.041^2.
Evaluating this equation, we get:
h ā 22.04 m.
Therefore, the ball will reach a height of approximately 22.04 meters.