a body is projected with speed 40root2 m/s downwards an angle of 45 degree from vertical if h=100 m find 1. time of flight 2. range
Asked by sohail.saba0311 | 9th Sep, 2021, 08:23: PM
As shown in above figure, an object is projected downward from a height h = 100 m above ground level
with velocity u 40√2 m/s so that angle of projection is 45o with vertical.
Projection velocity is resolved into two components,
(i) horizontal component ux = 40 m/s and (ii) vertical component uy = 40 m/s
Time of flight t , i.e., time taken for the object to reach ground is determined from the following equation
h = uy t + [ (1/2) g t2 ] .................... (1)
where g is acceleration due to gravity .
By substituting values in eqn.(1) , we get the following quadratic equation
100 = 40 t + 4.9 t2 or 4.9 t2 + 40 t -100 = 0
By solving above quadratic equation, we get t = 2 s
Range R , i.e. Horizontal distance from projection point is determined by following equation
R = ux t
where t is time of flight
Hence, we get , R = 40 × 2 = 80 m
Answered by Thiyagarajan K | 9th Sep, 2021, 09:59: PM
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