A small ball is thrown horizontally towards a vertical wall 1.2m away. It hits the wall 0.8m below its initial horizontal level. At what speed does the small ball hit the wall?(Neglect the air resistance.)
Please give me steps.thx!
physics
2009-12-30 5:14 am
回答 (2)
2009-12-30 5:35 am
✔ 最佳答案
Use equation of motion to calculate the vertical velocity component.using v^2 = u^2 + 2a.s
with u = 0 m/s, a = -g (=-10 m/s2), v =?, s = -0.8 m
hence, v^2 = 2(-10)(-0.8) (m/s)^2
v = 4 m/s
The time of travel of the ball is calculated by the equation of motion
s = (1/2)(u+v).t
hence t = 2 x 0.8/4 s =0.4 s
The horizontal velocity component of the ball = 1.2/0.4 s = 3 m/s
Therefore, speed of ball = square-root[3^2 + 4^2] m/s = 5 m/s
2009-12-30 6:40 am
The verticle speed of the ball hitting the wall:
V(y)^2=2gy V(y)=(2gy)^(1/2)=(2X10X0.8)^(1/2)=16^(1/2)=4m/s
y=(1/2)gt^2 then t^2=2y/g=2x0.8/10=0.16 t=0.4s
The horizontal speed of the ball hitting the wall:
V(x)=X/t=1.2/0.4=3m/s
The speed of the ball hitting the wall:
V={4^2+3^2}^1/2=25^(1/2)=5m/s
V(y)^2=2gy V(y)=(2gy)^(1/2)=(2X10X0.8)^(1/2)=16^(1/2)=4m/s
y=(1/2)gt^2 then t^2=2y/g=2x0.8/10=0.16 t=0.4s
The horizontal speed of the ball hitting the wall:
V(x)=X/t=1.2/0.4=3m/s
The speed of the ball hitting the wall:
V={4^2+3^2}^1/2=25^(1/2)=5m/s
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