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A ball is thrown vertically upwards with a velocity of 49 m/s.... - Science
A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate
(i) the maximum height to which it rises,
(ii) the total time it takes to return to the surface of the earth.
Given:
Initial velocity (
u) =49 \text{ m/s}Final velocity at maximum height (
v) =0 \text{ m/s}Acceleration due to gravity (
g) =-9.8 \text{ m/s}^2(upward motion)
(i) Maximum height (h):
Using the equation: v^2 - u^2 = 2gh0^2 - (49)^2 = 2 \times (-9.8) \times h-2401 = -19.6hh = \frac{2401}{19.6}h = 122.5 \text{ m}
(ii) Total time (t):
First, find the time to reach the top using: v = u + gt0 = 49 + (-9.8)t9.8t = 49t = \frac{49}{9.8} = 5 \text{ s}
Total time = Time of ascent + Time of descent
Since time of ascent = time of descent:
Total time =5 + 5
Total time = 10 \text{ s}