A cricket ball is thrown at a speed of 30ms^-1 in a direction 30 degrees above the horizontal. Calculate:
1. The maximum height.
2. The time taken to return to the same level.
3. The distance from the thrower to where the ball returns to the same level.
I would assume g as being 10ms^-2 therefore for the maximum height, I got: h = 1/2g (v sinθ)^2 =0.5×10^-2(30sin(30))^2 =1.125
Is this correct?
How would I do the other two please?
I don't understand any of this because I don't get how you would substitute the values into the equation?
Here,velocity of projection,u=30 m/s
Angle of projection=30°
We have,maximum height attained by a projectile is
H=u^2 sin tita/2g (u square sin tita divided by 2g)
=30^2×sin 30/2×9.8
=900×1/2/(whole divided by)19.6
=450/19.6
=22.95
=23 m (approx)
To calculate these values, we can use the equations of motion. Here's how you can find the answers step by step:
1. The maximum height:
- The motion of the cricket ball can be divided into two parts: vertical motion and horizontal motion.
- The initial vertical velocity can be found by decomposing the given speed of the ball. The vertical component can be calculated as v₀y = v₀ * sin(θ), where v₀ is the initial speed (30 m/s) and θ is the angle above the horizontal (30 degrees). So, v₀y = 30 m/s * sin(30°).
- The maximum height occurs when the vertical velocity becomes zero. At the highest point, the vertical component of velocity is zero and only the force of gravity acts upon the ball.
- The equation to calculate the maximum height (h) is h = (v₀y)² / (2 * g), where g is the acceleration due to gravity (approximately 9.8 m/s²).
- Substitute the values we found into the equation to solve for h.
2. The time taken to return to the same level:
- Since the vertical motion is symmetrical, the time taken to reach the highest point is equal to the time taken to descend and return to the same level.
- The equation to calculate the time of flight (T) is T = 2 * (v₀y) / g, where v₀y is the initial vertical velocity and g is the acceleration due to gravity.
- Substitute the values we found into the equation to solve for T.
3. The distance from the thrower to where the ball returns to the same level:
- The horizontal distance can be calculated using the equation d = v₀x * T, where v₀x is the initial horizontal velocity and T is the time of flight.
- The initial horizontal velocity can be found by decomposing the given speed of the ball. The horizontal component can be calculated as v₀x = v₀ * cos(θ), where v₀ is the initial speed (30 m/s) and θ is the angle above the horizontal (30 degrees).
- Substitute the values we found into the equation to solve for d.
Now, you have the step-by-step guidance to find the required values. Plug in the given values to find the answers.
h = 1/2g (v sinθ)^2
v sinθ t - 4.9t^2 = 0
r = v^2/g sin2θ