A ball is shot from the top of a building with an initial velocity of 19m/s at an angle theta = 45 above the horizontal.
Vxo and Vyo, are 13.4m/s
If a nearby building is the same height and 50m away, how far below the top of the building will the ball strike the nearby building?
I found the time from x=vt, 50=13.4*t
t= 3.73s
then y= yo + Vyo*t - 0.5*at^2
y= yo + 13.4*3.73 - 0.5*9.8*(3.73)^2..
shouldn' t the answer with that part of formula which is " Vyo*t - 0.5*at^2 " give me how far it could be?
ty :)
Well, it seems like you're calculating the vertical displacement using the equation y = yo + Vyo*t - 0.5*a*t^2. However, that equation gives you the height of the ball at a certain time, not the horizontal distance it will travel.
To find out how far below the top of the building the ball will strike the nearby building, you need to calculate the horizontal displacement. Since the nearby building is 50m away, and the ball has an initial horizontal velocity of 13.4 m/s, you can use the equation x = vxt to find the horizontal displacement.
Using x = vxt, you have x = 13.4 m/s * 3.73 s = 49.942 m.
So, the ball will strike the nearby building approximately 49.942 meters below the top of the building.
No, the formula "Vyo*t - 0.5*at^2" does not give you the distance. It gives you the vertical displacement (or height) of the ball at a given time.
To find the distance the ball travels horizontally, you need to use the horizontal component of the initial velocity (Vxo) and the time of flight (t).
The horizontal distance traveled by the ball can be calculated using the formula:
distance = Vxo * t
Substituting the values, distance = 13.4 m/s * 3.73 s, the horizontal distance traveled by the ball is approximately 49.96 m (or 50 m considering rounding).
Therefore, the ball will strike the nearby building approximately 50 m away horizontally.
The formula "Vyo*t - 0.5*at^2" gives you the vertical distance traveled by the ball. However, since we are interested in finding the horizontal distance, we need to consider the horizontal component of the ball's motion.
To find the horizontal distance, we first need to find the horizontal velocity (Vxo) of the ball, which is given as 13.4 m/s.
Since the ball is in projectile motion, the horizontal distance it travels is equal to the horizontal velocity multiplied by the time of flight (t). In this case, the time of flight is 3.73 seconds (as you calculated). So,
Horizontal distance = Vxo * t
Horizontal distance = 13.4 m/s * 3.73 s
Horizontal distance = 49.982 m (approximately)
Therefore, the horizontal distance the ball will travel is approximately 49.982 meters.
Now, to find the vertical distance below the top of the nearby building where the ball will strike, we need to calculate the vertical displacement (y) using the equation:
y = yo + Vyo*t - 0.5*a*t^2
Here, yo is the initial vertical position (which is the height of the building), Vyo is the initial vertical velocity, a is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time of flight.
Since the ball is shot from the top of the building, yo would be equal to the height of the building. If the building is 50 meters tall, then yo = 50 meters.
Now, substituting the values into the equation:
y = 50 m + 13.4 m/s * 3.73 s - 0.5 * 9.8 m/s^2 * (3.73 s)^2
Calculating this equation will give you the vertical distance below the top of the building where the ball will strike the nearby building.
L=vₒ²•sin2α/g =19²sin2•45/9.8=36.8 m
At the distance of 36.8 m from the first building the ball is at the level of starting point and has the velocity v(x)=v₀•cos45° =13.4 m/s
The distance to the second building is 50-36.8 = 13.2 m
this distance is covered for t=l/v(x) = 13.2/13.4 =0.99 s.
h= gt²/2=9.8•0.99²/2=4.8 m.