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?
a) No. Write an equation for the vertical direction only. The vertical component of the initial velocity is zero.
y = 100 - (g/2)t^2
Set y = 0 ands solve for t.
b)The initial value of x is zero and the horizontal component of velocity remains 30 m/s. Gravity has no hoorizontal component
c) No. X = Xo + Vt will not tell you the horizontal componsnt of velocity. After solving a) for t, calculate Vy = -gt. Vx remains the initial value
x = 30
d) yes, your method is correct
Yes, you are on the right track with the equations. Let's go through each part of the question and see how to use these equations to solve them:
a) To determine the time necessary for the projectile to reach the ground, you would indeed use the equation you provided: y = y_0 + v_0t + 1/2 at^2. In this case, since the projectile is shot horizontally, the initial vertical velocity (v_0) is 0. The initial height (y_0) is given as 100m, and the acceleration (a) is the acceleration due to gravity, which is -10m/s^2 in the negative vertical direction. Therefore, you would have the equation: 100 = 0 + 0t + 1/2(-10)t^2. Solve this equation for t to find 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 = x_0 + v_xt. In this case, the initial horizontal velocity (v_x) is given as 30m/s. The initial horizontal position (x_0) is 0, as the projectile is shot from the roof of the building. The time (t) can be found from part (a). Plug in these values into the equation to calculate the horizontal distance.
c) You are correct with the equations for the horizontal and vertical components of velocity just before the projectile hits the ground. The horizontal component of velocity (vx) remains constant throughout the motion, as there is no horizontal acceleration. Therefore, vx = v_x. The vertical component of velocity (vy) changes due to the acceleration of gravity. You can use the equation vy = v_0 + at, where v_0 is the initial vertical velocity (0 in this case) and a is the acceleration due to gravity (-10m/s^2).
d) To find 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), where vx and vy are the horizontal and vertical components of velocity, respectively. Plug in the values of vx and vy calculated in part (c) to find the magnitude of the total velocity.
Remember to take note of the signs and directions while plugging in values and solving equations, as positive and negative directions of velocity and displacement are important in this problem.