A ball is thrown at an angle of 30 degrees from a building of height 20m , calculate the time taken to reach point p
Answers
Let time taken to reach maximum height be t from top of tower
V = U + at
0=usinα – gt
t = usinα/g = 2sec
therefore total time during projectile T1 = 2 t = 4sec
now when particle returned to the same height it have same velocity usinα but in downward direction ( as shown in fig)
now apply
S=usinα * t2 +1/2 at22
156.8 = 39.2/2 * t2 + ½ * 9.8 * t22
Divide both side by 9.8
16 = 2 t2 + ½ t22
t22 +4 t2 – 32 = 0
0n solving the quadratic equation we get t2 = 4sec
Therefore total time take during whole journey = (T1 + t2) = 4+4 = 8sec
T=8sec
Now for distance travelled in x direction = ucosα(T1) + ucosα( t2)
Since the horizontal velocity will be same throughout the motion
Therefore total distance in x direction = ucosα(T1 + t2) = (39.2*√3/2 )*(4+4)
= 33.94 * 8
= 271.57m
Total distance travelled =271.57m
Answer:
Throw ball at velocity (v, 30°) from 20 m tower. Reach ground 100 m from tower base after total flight time t. Find t.
T = time to return to launch height 20 m, with negative of vertical part of launch velocity:
v*sin(30) = gT/2
v = gT
v/g = T time to rise & return to 20 m.
t = T + (time to reach ground after T).
v*cos(30)t = 100 Total horizontal travel.
vt = 100/0.866 = 115.47
We need to eliminate v and T...
vt= gTt
gTt = 115.47
T = 115.47/(gt) = 11.77/t
v = gT = 115.47/t
vT = gT^2 = 1359/t^2
Fall from height of tower: initial vertical velocity -0.5v, time t-T, distance -20 m:
-20 = -0.5v(t-T) -0.5g(t-T)^2
Flip signs and expand square expression:
20 = 0.5v(t-T) + 0.5g(t^2 -2Tt + T^2)
Substitute gTt = 115.47:
20 = 0.5v(t-T) + 0.5g(t^2 + T^2) -115.47
Substitute vt and combine constants:
20 = -57.7 -0.5vT +0.5gt^2 +0.5gT^2
Substitute vT and T:
20 = -57.7 -679.5/t^2 +4.9t^2 +679.5/t^2
Add 57.7 both sides; sum 1/t^2 components:
77.7 = 4.9t^2
t = sqrt(77.7/4.9). Flight time 3.98 seconds