A projectile is shot horizontally at 30m/s from the roof of a building 100m tall. Use g=10m/s^2. Choose + vertical direction to be pwards.
a) determine the time necessary fot the projectile to reach the ground below
b) determine the horizontal distance from the base of the building that the projectile lands
c) determine the horizontal and vertical components of the velocity just before the projectile hits the ground
d) determine the magnitude of the total velocity just before the projectile hits the ground.
a)would I use y=y_0 +v_0t +1/2 at^2
100=0+30t+1/2(10)t^2 and solve for t?
b) would i use x=x_0+v_xt
c)for the horizontal componet i believe i use v=v_0 +at and for the vertical: x=x_0+vt
d) v_total =sqrt(vx^2 +vy^2)
Are these equations I came up with correct to solve this problem?
Calculate how far a boy standing on a 5m high wall will be able to shoot a stone with a projecting device if the stone is projected upwards 45 and it has an initial velocity of 20m/s
Yes, the equations you have written are correct and can be used to solve the problem. Let's go through each part step by step:
a) To determine the time necessary for the projectile to reach the ground, you can use the equation:
y = y0 + v0t + (1/2)at^2
Here, y is the final position (in this case, 0 since it reaches the ground), y0 is the initial position (100m), v0 is the initial velocity (30m/s), a is the acceleration due to gravity (-10m/s^2 since it acts downward), and t is the time.
Plugging in the values, you will get:
0 = 100 + 30t + (1/2)(-10)t^2
Solving this quadratic equation will give you the time it takes for the projectile to reach the ground.
b) To determine the horizontal distance from the base of the building that the projectile lands, you can use the equation:
x = x0 + vxt
Here, x is the horizontal distance, x0 is the initial horizontal position (0 since it starts from the base of the building), vx is the constant horizontal velocity (since there is no acceleration horizontally), and t is the time.
Plugging in the values, you will get:
x = 0 + 30t
Solving this equation will give you the horizontal distance.
c) Just before the projectile hits the ground, its vertical velocity will be the same as its initial vertical velocity, and its horizontal velocity will remain constant.
For the horizontal component, you are correct in using the equation:
v = v0 + at
Here, v is the final horizontal velocity, v0 is the initial horizontal velocity (30m/s), a is the horizontal acceleration (0 since there is no horizontal acceleration), and t is the time.
Plugging in the values, you will get the horizontal component of velocity just before hitting the ground.
For the vertical component, you can use the equation:
y = y0 + vt
Here, y is the final vertical position (0 since it reaches the ground), y0 is the initial vertical position (100m), v is the final vertical velocity, and t is the time.
Since the projectile is shot horizontally, there is no vertical acceleration, so the initial vertical velocity remains constant throughout the motion.
Plugging in the values, you will get the vertical component of velocity just before hitting the ground.
d) To determine the magnitude of the total velocity just before the projectile hits the ground, you can use the equation:
v_total = sqrt(vx^2 + vy^2)
Here, vx is the horizontal component of velocity and vy is the vertical component of velocity, both just before hitting the ground.
Plugging in the values, you will get the magnitude of the total velocity.
Remember to check units and signs while solving the equations.