A ball is thrown horizontally from a building 63.0 m high with a speed of 25.0 m/s. Find the (a)vertical and horizontal components of the ball's initial velocity, (b) time flight, (c) distance from the foot of the building where the ball will strike the ground, and (d) velocity when the ball will strike the ground.
+36916 The ball is thrown HORIZONTALLY, so the initial vertical velocity = 0 m/s and the initial horizontal velocity = 25 m/s.
1 The ball's 'x' position is given by x = x0 + v0t x0=0 vo=25 (there is no 'a' in the x direction)
2 The ball's 'y' position is y = yo + vot + 1/2 a t^2 where yo=63 vo=0 a = -9.8 m/s^2
Use 2 to find t when y = 0 (when the ball hits the ground)
t = 3.59 s
Use this time to find the x distance the ball travels before hitting the ground (equation 1)
x = 25(3.59) = 89.64 m
3 Velocity when it hits the ground will be BOTH an x velocity and a y velocity.
the x velocity is given as 25 m/s
the y velocity is given by vy = vo + at v0= 0 a = -9.8 m/s^2 t = 3.59 s
vy = -9.8(3.59) = -35.18 m/s
'VECTORIALLY' add x and y together to get the velocity MAGNITUDE sqrt(25^2 + 35.18^2)= 43.16 m/s
+36916 The ball is thrown HORIZONTALLY, so the initial vertical velocity = 0 m/s and the initial horizontal velocity = 25 m/s.
1 The ball's 'x' position is given by x = x0 + v0t x0=0 vo=25 (there is no 'a' in the x direction)
2 The ball's 'y' position is y = yo + vot + 1/2 a t^2 where yo=63 vo=0 a = -9.8 m/s^2
Use 2 to find t when y = 0 (when the ball hits the ground)
t = 3.59 s
Use this time to find the x distance the ball travels before hitting the ground (equation 1)
x = 25(3.59) = 89.64 m
3 Velocity when it hits the ground will be BOTH an x velocity and a y velocity.
the x velocity is given as 25 m/s
the y velocity is given by vy = vo + at v0= 0 a = -9.8 m/s^2 t = 3.59 s
vy = -9.8(3.59) = -35.18 m/s
'VECTORIALLY' add x and y together to get the velocity MAGNITUDE sqrt(25^2 + 35.18^2)= 43.16 m/s
