Aston is thrown upwards with an initial velocity of 25 m/s at an angle of 30 degree to the ground, a,calculate the vertical component of the velocity, b fine the maximum height of the stone ,c how long will it take for the stone reach the maximum height D, how long it take between the stone was thrown and when it come back to ground
a. The vertical component of the velocity is 12.5 m/s.
b. The maximum height of the stone is 93.75 m.
c. It will take 3.75 seconds for the stone to reach its maximum height.
d. It will take 7.5 seconds for the stone to come back to the ground.
To solve this problem, we'll use the equations of motion for projectile motion.
a) The vertical component of the initial velocity can be found using the equation:
v_y = v * sin(θ)
where v is the initial velocity and θ is the angle with the ground.
Plugging in the values, we get:
v_y = 25 m/s * sin(30°)
v_y ≈ 12.5 m/s
b) To find the maximum height of the stone, we can use the equation:
h = (v_y^2) / (2 * g)
where v_y is the vertical component of the velocity and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values, we get:
h = (12.5 m/s)^2 / (2 * 9.8 m/s^2)
h ≈ 15.95 meters
c) The time taken for the stone to reach the maximum height can be found using the equation:
t = v_y / g
Plugging in the values, we get:
t = 12.5 m/s / 9.8 m/s^2
t ≈ 1.28 seconds
d) The total time taken for the stone to return to the ground can be found by doubling the time taken to reach the maximum height:
Total time = 2 * t
Plugging in the value of t, we get:
Total time = 2 * 1.28 seconds
Total time ≈ 2.56 seconds
So, the answers are:
a) The vertical component of the velocity is approximately 12.5 m/s.
b) The maximum height of the stone is approximately 15.95 meters.
c) It takes approximately 1.28 seconds for the stone to reach the maximum height.
d) It takes approximately 2.56 seconds for the stone to return to the ground.