HELP WITH AS-LEVEL PHYSICS QUETION

[PinkRhinos]

Hello so ,even though my friend told me it is simple ,they cannot explain to me what the answer is.

thanks guys.

"A boy is throwing a ball over a wall. He is standing 5m away from the wall,whose top is 3m highr thzn his hand at the instant he releases the ball." "Show that,if the ball is thrown with a velocity of 11m/s at an angle of 45° to the horizontal,the ball will just go over the wall.

 

[geek101]

Hello so ,even though my friend told me it is simple ,they cannot explain to me what the answer is.

thanks guys.

"A boy is throwing a ball over a wall. He is standing 5m away from the wall,whose top is 3m highr thzn his hand at the instant he releases the ball." "Show that,if the ball is thrown with a velocity of 11m/s at an angle of 45° to the horizontal,the ball will just go over the wall.

draw a right angle triangle.. with 3 being the vertical length, and 5 the horizontal. This is because he is standing 5 m away...so itll be the horizontal distance..and the 3 m high then the point of projection.
use Pythagoras theorem to calculate the length of the hyp (x)
x^2 = 3^2 + 5^2
x = 34 ^0.5 m

at the point where it is right on top of the wall, the v = 0 ms-1
and we want the angular velocity, which will be the initial velocity so, u = ? ms-1
we have the approximate distance the ball will travel = 34 ^ 0.5 m
and a = -9.81 ms-2
use the kinematic equation : V^2 = U^2 + 2AS
substitute and ull get u = 10.69... which is approximately 11 ms-1

for the second part...im getting the angle as 40, are you sure its 45

 

[PinkRhinos]

draw a right angle triangle.. with 3 being the vertical length, and 5 the horizontal. This is because he is standing 5 m away...so itll be the horizontal distance..and the 3 m high then the point of projection.
use Pythagoras theorem to calculate the length of the hyp (x)
x^2 = 3^2 + 5^2
x = 34 ^0.5 m

at the point where it is right on top of the wall, the v = 0 ms-1
and we want the angular velocity, which will be the initial velocity so, u = ? ms-1
we have the approximate distance the ball will travel = 34 ^ 0.5 m
and a = -9.81 ms-2
use the kinematic equation : V^2 = U^2 + 2AS
substitute and ull get u = 10.69... which is approximately 11 ms-1

for the second part...im getting the angle as 40, are you sure its 45

it IS 45 degrees, thanks alot for the reply, i will look at it when i get back from school