Explanation:
Let us first calculate long does it take to go 12m at 30m/s( assumed speed)
12/30 = 0.4 seconds
horizontal distance the ball drop in that time
H= (0)(0.4)+1/2(-9.8)(0.4)2
H= -0.78m
negative sign shows that the height of the ball at the net from the top.
Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m
As 1.62m>0.9m so the ball will clear the net.
H_1= V0y t’ + ½ g t’^2
-2.4= (0)t’ + ½ (-9.8) t’^2
t’= 0.69s
X’=V0x t’
X’=(30)(0.96)
X’= 20.7m
The acceleration due to gravity on Earth is 9.8 m/s per second.
Potential energy U = mgh
Given h = 123 m,
mg = F = 780 N
Then
U = (123)(780)
= 95940
= 9.59 x 10^4
Answer:
a) t = 1.6 s
b) d = 4.9 m
c) v = 16 m/s
d) θ = 79°
Explanation:
time of fall
t = √(2h/g) = √(2(12)/9.8) = 1.5649... s
d = vt = 3.1(1.56) = 4.8512...
vertical velocity vy = at = 9.8(1.56) = 15.336... m/s
v = √(15.336² + 3.1²) = 15.6464... m/s
θ = arctan(15.336/3.1) = 78.5724...°