A ball is thrown down from the top of a tower with a speed of 2.5 m/s. It hits the ground in 8 seconds. What is the height of the tower ? With what speed will the stone hit the ground ? Take g = 9.8 m/s2. [ Ans. (a) 333.6 m (b) 80.9 m/s ]
Answers
Answer:
Explanation:
The height of tower and horizontal range of the stone are 46.4 m and 55.4 m.
Explanation:
Given that,
Velocity = 16 m/s
Time t 4 sec
Angle = 30°
We need to calculate the height of the tower
Using equation of projectile 1
Sy = Uyt
29t²
+ ho
Where, ho = height
u =vertical velocity = u sin
t = time
Put the value in the equation
1 0 = 16 x x 4 1 2 x 9.8 x 16 + ho
ho 46.4 m
Now, The horizontal range of the stone is Using formula of horizontal range
Sx = Uxt
S = 16 cos 30° x 4
Sx = 16 X √3 2 x 4m
Sx 55.4 m
Hence, The height of tower and horizontal range of the stone are 46.4 m and 55.4 m.
=> Initial velocity, u = 0m/s
=> Acceleration due to gravity, g = 9.8 m/s²
=> Time taken to reach the ground, t = 2.5 m/s
=> Height, h = ?
Using equation,
=> s = u x t + 1/2gt²
=> s = 0 x 2.5 + 1/2 x 9.8 x 2.5 x 2.5
=> s = 0 x 4.9 x 2.5 x 2.5
=> s = 30.625 m
Therefore answer will be 30.625 meter.