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.
Posted this yesterday but still don't understand
Read the excellent article at wikipedia:
https://en.wikipedia.org/wiki/Trajectory
You can then understand why
max height = (v sin^2θ)/2g
range = (v^2 sin2θ)/2g
time in air solves
(v sinθ) t - g/2 t^2 = 0
To solve this problem, we can use the equations of motion for projectiles. Let's break it down step-by-step:
1. First, let's find the vertical component of the initial velocity. We can use the formula:
Vy = V * sin(θ)
where Vy is the vertical component of the velocity, V is the initial velocity, and θ is the angle above the horizontal.
Given V = 30 m/s and θ = 30 degrees, we can calculate Vy:
Vy = 30 * sin(30)
= 15 m/s
2. Next, let's find the time taken to reach the maximum height. We can use the formula:
t = Vy / g
where t is the time, Vy is the vertical component of the velocity, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in Vy = 15 m/s, we can find t:
t = 15 / 9.8
≈ 1.53 seconds
3. To find the maximum height, we can use the following formula:
H = (Vy^2) / (2 * g)
where H is the maximum height.
Plugging in Vy = 15 m/s and g = 9.8 m/s^2, we can calculate H:
H = (15^2) / (2 * 9.8)
= 22.96 meters
Therefore, the maximum height is approximately 22.96 meters.
4. Next, let's calculate the time taken for the ball to return to the same level. Since the time taken to rise is equal to the time taken to fall, we just need to double the time taken to reach the maximum height:
Total time = 2 * t
= 2 * 1.53
= 3.06 seconds
Therefore, the time taken to return to the same level is approximately 3.06 seconds.
5. Finally, let's find the horizontal distance the ball travels before returning to the same level. We can use the formula:
Range = Vx * t
where Range is the horizontal distance, Vx is the horizontal component of the velocity, and t is the time.
The horizontal component of velocity can be calculated using the formula:
Vx = V * cos(θ)
Given V = 30 m/s and θ = 30 degrees, we can find Vx:
Vx = 30 * cos(30)
≈ 25.98 m/s
Plugging in Vx = 25.98 m/s and t = 3.06 seconds, we can calculate the range:
Range = 25.98 * 3.06
≈ 79.51 meters
Therefore, the distance from the thrower to where the ball returns to the same level is approximately 79.51 meters.
To calculate the maximum height, time taken to return to the same level, and the distance from the thrower to where the ball returns to the same level, we can use projectile motion equations. Here's how you can approach each part:
1. Maximum Height:
To find the maximum height, we need to determine the vertical component of the initial velocity. Given that the ball is thrown at a speed of 30 m/s and at an angle of 30 degrees above the horizontal, we can calculate the vertical component of the velocity (Vy) using the equation:
Vy = V * sin(θ)
where V is the speed of the ball and θ is the angle. Substituting the values, we have:
Vy = 30 m/s * sin(30 degrees)
Vy = 15 m/s
Now, we can use the equation for maximum height (Hmax) in projectile motion:
Hmax = (Vy^2) / (2 * g)
where g is the acceleration due to gravity (approximately 9.8 m/s^2). Plugging in the values:
Hmax = (15 m/s)^2 / (2 * 9.8 m/s^2)
Hmax ≈ 11.47 m
The maximum height reached by the ball is approximately 11.47 meters.
2. Time Taken to Return to the Same Level:
To find the time taken to return to the same level, we can use the equation for the total flight time (T) in projectile motion:
T = (2 * Vy) / g
Plugging in the values:
T = (2 * 15 m/s) / 9.8 m/s^2
T ≈ 3.06 seconds
The time taken for the ball to return to the same level is approximately 3.06 seconds.
3. Distance from the Thrower to Where the Ball Returns to the Same Level:
To find the horizontal distance traveled by the ball, we need to determine the horizontal component of the initial velocity. We can calculate this using the equation:
Vx = V * cos(θ)
where V is the speed of the ball and θ is the angle. Substituting the values:
Vx = 30 m/s * cos(30 degrees)
Vx ≈ 25.98 m/s
Now, we can calculate the distance (D) using the equation:
D = Vx * T
Plugging in the values:
D = 25.98 m/s * 3.06 s
D ≈ 79.57 meters
The distance from the thrower to where the ball returns to the same level is approximately 79.57 meters.
I hope this explanation helps you understand how to solve the problem. If you have any further questions, feel free to ask!