In a hockey match a hockey ball is hit 18m from the front of the goal. The ball leaves the hockey stick with initial velocity v at angle theta to the horizontal ground. The ball passes over the goal at a maximum height of 2.0m.
The initial vertical component of the velocity of the ball is 6.3ms-1. Air resistance has negligible effect on the motion of the ball.
Show that the time take for t to reach maximum height is about 0.6s.
I used suvat and got
s-2
u-6.3
no v
a-9.81
t ??????
so s=ut + 1/2at^2 has to be used to find t. I kept trying to rearrange find the answer but couldn’t do please help using suvat!!!
Show that the time take for t to reach maximum height is about 0.6s.
I am calling initial speed s, not v. For me v is vertical component
vertical problem only
Vi = s sin theta
v = Vi - g t = Vi - 9.8 t
at top v = 0
0 = Vi - 9.8 t
so
9.8 t = s sin theta
t = (s/9.8) sin theta at top
now height h
h = 0 + Vi t -4.9 t^2
2 = Vi t - 4.9 t^2
4.9 t^2 - Vi t + 2 = 0
but we know Vi = 9.8 t
4.9 t^2 - 9.8 t^2 + 2 = 0
4.9 t^2 = 2
t = 0.64 seconds to top
now you can go back and find Vi which better be 6.3 and theta and u = 18/t and then tan theta = Vi/u
Sure, let's use the suvat equations to solve the problem.
We know that the initial vertical component of the velocity (u) is 6.3 m/s. The acceleration (a) is -9.81 m/s^2 (taking upward as positive and the acceleration due to gravity). The maximum height (s) is 2.0 m.
Let's use the equation s = ut + (1/2)at^2 to find the time taken to reach the maximum height.
By substituting the known values into the equation, we get:
2.0 = (6.3)t + (1/2)(-9.81)t^2
Simplifying the equation, we have:
0 = -4.905t^2 + 6.3t - 2
To solve this quadratic equation, we can either factorize it or use the quadratic formula. However, let's take a different approach and use a little bit of humor to make it interesting.
Why did the ball go to therapy?
Because it had too many complex issues to solve!
Okay, let's get back to solving it.
Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / (2a), where a = -4.905, b = 6.3, and c = -2.
Plugging in the values, we have:
t = (-(6.3) ± √((6.3)^2 - 4(-4.905)(-2))) / (2(-4.905))
t = (-6.3 ± √(39.69 + 39.24)) / (-9.81)
t = (-6.3 ± √(78.93)) / (-9.81)
Here comes the punchline.
Why did the square root go to the party?
Because it always knows how to get to the root of the fun!
Alright, let's continue.
t = (-6.3 ± 8.888) / (-9.81)
Using the positive form of the square root:
t = (-6.3 + 8.888) / (-9.81)
t = 2.588 / (-9.81)
t ≈ -0.26 s
Since time cannot be negative in this context, we discard this solution.
Now using the negative form:
t = (-6.3 - 8.888) / (-9.81)
t = -15.188 / (-9.81)
t ≈ 1.55 s
Although this is the correct time taken, it refers to the total time of flight. To find the time taken to reach the maximum height, we divide it by 2:
t = 1.55 / 2
t ≈ 0.775 s
To be a little more accurate, let's round it to one decimal place:
t ≈ 0.8 s
And there you have it! The time taken for the ball to reach maximum height is approximately 0.8 seconds. I hope I was able to help you with a bit of humor along the way!
To find the time taken for the ball to reach its maximum height, we can use the equation for vertical motion:
s = ut + (1/2)at^2
Given:
s = 2.0 m (maximum height)
u = 6.3 m/s (initial vertical component of velocity)
a = -9.81 m/s^2 (acceleration due to gravity)
Substituting these values into the equation, we have:
2.0 = (6.3)t + (1/2)(-9.81)(t^2)
Rearranging the equation to solve for t:
0 = (1/2)(-9.81)(t^2) + (6.3)t - 2.0
Multiplying both sides of the equation by 2 to eliminate the fraction:
0 = -9.81(t^2) + 12.6t - 4.0
Now we have a quadratic equation in the form of: at^2 + bt + c = 0, where a = -9.81, b = 12.6, and c = -4.0.
We can solve for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values of a, b, and c into the formula:
t = (-(12.6) ± √((12.6)^2 - 4(-9.81)(-4.0))) / (2(-9.81))
Simplifying the equation:
t = (-12.6 ± √(158.76 - 156.96)) / (-19.62)
t = (-12.6 ± √(1.8)) / (-19.62)
t = (-12.6 ± 1.34) / (-19.62)
There are two possible solutions for t, which are:
t1 = (-12.6 + 1.34) / (-19.62) ≈ 0.594 s
t2 = (-12.6 - 1.34) / (-19.62) ≈ -0.989 s
Since time cannot be negative in this context, we can discard t2 as the extraneous solution.
Therefore, the time taken for the ball to reach its maximum height is approximately 0.6 seconds (or 0.594 seconds to be more precise).
To find the time it takes for the ball to reach its maximum height, we can use the equation of motion:
s = ut + (1/2)at^2
Given:
s = 2.0m (maximum height)
u = 6.3 m/s (initial vertical component of velocity)
a = -9.81 m/s^2 (acceleration due to gravity)
Substituting these values into the equation, we get:
2.0 = 6.3t + (1/2)(-9.81)t^2
Simplifying, we have:
2.0 = 6.3t - 4.905t^2
Rearranging the equation to form a quadratic equation:
4.905t^2 - 6.3t + 2.0 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Where a, b, and c are the coefficients of the quadratic equation. In our case, a = 4.905, b = -6.3, c = 2.0.
Plugging in these values into the quadratic formula, we can calculate t:
t = (-(-6.3) ± √((-6.3)^2 - 4(4.905)(2.0))) / (2(4.905))
Simplifying further, we have:
t = (6.3 ± √(39.69 - 39.24)) / 9.81
t = (6.3 ± √0.45) / 9.81
Since we're interested in the time it takes for the ball to reach its maximum height, we need to consider the positive value in the square root:
t = (6.3 + 0.671) / 9.81
t ≈ 0.971 / 9.81
t ≈ 0.099 seconds
Therefore, the time it takes for the ball to reach its maximum height is approximately 0.099 seconds or approximately 0.1 seconds (rounded to one decimal place).