Science
A batsman hits a cricket ball from ground level. It is caught 1.8 seconds later by the third man standing 43.4 m from the bat. The ball leaves the bat at a speed of 26m/s. Calculate how high above the ground it is caught
Your School SUBJECT is science.
the x-velocity is 43.4/1.8 = 24.111 m/s
so, the ball is hit at an angle θ such that cosθ = 24.111/26, so θ = 21.97°
So, the y-velocity is 26sin21.97° = 9.73 m/s.
The height of the ball is
y = 9.73t - 4.9t^2
So, when caught after 1.8 seconds,
y(1.8) = 1.638
To calculate the height above the ground at which the ball is caught, we need to consider the vertical motion of the ball. We can use the equations of motion to solve this problem.
First, let's analyze the vertical motion of the ball. We can assume that the only force acting on the ball during its flight is gravity, which causes it to accelerate downward at a rate of 9.8 m/s².
We know that the initial vertical velocity of the ball is 0 m/s because it is hit from ground level. The final vertical velocity of the ball at the moment it is caught will also be 0 m/s because the ball reaches its maximum height and starts falling downward.
Using the equation of motion for vertical displacement:
Δy = v₀t + (1/2)at²
Where:
Δy is the vertical displacement (the height above the ground)
v₀ is the initial velocity in the vertical direction (0 m/s)
t is the time taken (1.8 seconds)
a is the acceleration in the vertical direction (-9.8 m/s²)
Substituting the known values into the equation, we get:
Δy = 0 + (1/2)(-9.8)(1.8)²
Simplifying the equation, we have:
Δy = -4.9 * (1.8)²
Calculating further:
Δy = -4.9 * 3.24
Δy = -15.876
Since the ball is caught above the ground, the displacement Δy is negative. To determine the height above the ground, we take the absolute value:
Height above ground = 15.876 meters
Therefore, the ball is caught at a height of approximately 15.876 meters above the ground.