a ball is thrown upward with speed 10m/s^1 from the top of the tower reaches the ground with a speed 20m/s^1 . the height if the tower is
Answers
V^2 = U^2 + 2AS
v = 20 m/s
u = 0 m/s (velocity at highest point)
It implies,
400 = 0+ (2 * 10 * s)
400 = 20s
s = 20m (total distance covered by rock while going down from its highest point)
V^2 = U^2 + 2AS
0 = (10^2) + (2*-10*s)
100 = 20s
s = 5 (distance covered by stone when thrown up with an intital velocity of 10m/s)
Therefore tower is 20-5 = 15m tall.
A ball is thrown upward with speed 10m/s from the top of the tower reaches the ground with a speed 20m/s.
We have to find the height of the tower.
From above data we have; initial velocity is 10 m/s and final velocity is 20 m/s.
First equation of motion:
v = u + at
Second equation of motion:
s = ut + 1/2 at²
Third equation of motion:
v² - u² = 2as
We have value of v and u. And we know that acceleration due to gravity is 9.8 approx. 10 m/s². So,
Using the Third Equation Of Motion:
v² - u² = 2as
Substitute the known values,
→ (20)² - (10)² = 2(10)s
→ 400 - 100 = 20s
→ 30 = 2s
Divide by 2 on both sides,
→ 30/2 = 2s/2
→ 15 = s
Therefore, the height of the tower is 15 m.