A ball thrown with a speed of 100ms^-1 attains a height of 150m. ( take g= 9.8ms^-2) calculate the: (a) time of flight (b) angle of projection (c) range?

Answer 1

Recall the basic equation of motion here. The height y is
y = 100sinθ t - 4.9t^2
and the ball's upward speed is
v(t) = 100sinθ - 9.8t
(b) The max height is achieved when the velocity becomes zero. That is, when
t = 100sinθ/9.8 = 10.2 sinθ
Now, we know that the max height is 150, so
100sinθ * 10.2 - 4.9*10.2^2 = 150
θ = 40°

(a) Now we know that our height is roughly
y = 64.7t - 4.9t^2
So, the time of flight is 13.2 seconds

(c) since the horizontal speed is a constant 100cosθ = 76.6 m/s, the range is
76.6 * 13.2 = 1011 m

Answer 2

Y^2 = Yo^2 + 2g*h = 0,
Yo^2 + (-19.6)150 = 0.
Yo = 54.2 m/s. = Ver. component of initial velocity.

b. Yo = Vo*sinA = 54.2,
100*sinA = 54.2,
A = 32.8o.

c. Range = Vo^2*sin(2A)/g = 100^2*sin(65.6)/9.8 = 929.3 m.

a. Range = Xo*T = 100*Cos32.8 * T = 929.3,
T = 11.1 s. = Time of flight.

Answer 3

I AM GREATFUL WITH YOUR SOLUTION THAT PROVIDED TO ME. THANKS A LOAD